Math 202 (Introduction to Real Analysis) Fall and Spring 2007-2008Lecture 1

Instructor: Dr. Kin Y. LiOffice: Room 3471 Office Phone: 2358 7420 e-mail address: makyli@ust.hkOffice Hours: Tu. 3:30pm - 4:30pm (or by appointments)

Prerequisite: A-level Math or One Variable Calculus

Website for Lecture Notes and Practice Exercises: http://www.math.ust.hk/�makyli/UG.html

Grade Scheme: Homeworks (5%), Tutorial Presentations (5%), Fall Midterm (15%), Fall Final (30%),Spring Midterm (15%), Spring Final (30%)

All records of grades will be put on the website http://grading.math.ust.hk/checkgrade/ as soon as they areavailable. At the end of Fall semester, students who have done a reasonable job on homeworks and exams will be givena pp-grade for permission to proceed to the next semester. Those who have not done a reasonable job in the opinionof the instructor (for example, received less than 40 marks in the fall midterm and final exam marks combined )will get a F-grade and will repeat the course next year. For those who are permitted to proceed to the second semester,the overall grade will be given at the end of the second semester. This course is essentially graded by curve withone exception, namely students who achieve 40% or less of the overall grade will fail the course.

Students should make copies of homeworks before submitting the originals. In case homeworks are notreceived, students will be required to resubmit copies within a short period of time (may be less than a day).

For tutorial presentations, students will form groups (of 1 to 3 students) and present solutions to assigned problemsin the tutorial sessions. All members of a group must attend the sessions to assist in answering possible questions frompresentations. Marks will be deducted for failure to present solutions or for absence in supporting his/her group.

Course Description: Math 202 is a 2 semester course. It is the first of two required courses on analysis for Mathmajors. It is to be followed by Math 301 (Real Analysis). This course will focus on the proofs of basic theoremsof analysis, as appeared in one variable calculus. Along the way to establish the proofs, many new concepts will beintroduced. These include countability, sequences and series of numbers, of functions, supremum/infimum, Cauchycondition, Riemann integrals and improper integrals, etc. Understanding them and their properties are important forthe development of the present and further courses.

Recommended Text:Steven Lay, Analysis With an Introduction to Proof , 4th ed., Prentice Hall, 2005.

References:

1. J. A. Fridy, Introductory Analysis; 2nd ed., Academic Press, 2000.2. Manfred Stoll, Introduction to Real Analysis, Addison Wesley, 1997.3. Walter Rudin, Principles of Mathematical Analysis, 3rd ed., McGraw-Hill, 1976.4. Tom Apostol, Mathematical Analysis, 2nd ed., Addison-Wesley, 1974.

*5. Chinese Solution Manual to Tom Apostol’s Mathematical Analysis, 2nd ed.

Reminders: Students are highly encouraged to come to office hours for consultation. This is a difficult course formany, but not all students. Although there are lecture notes, students should attend all lectures and tutorials aslectures notes are only brief records of materials covered in class, which may contain typographical errors. Of course,questions from students and answers from instructors or other digressions will not be recorded. Students are advisedto take your own notes. All materials presented in lectures and tutorials as well as proper class conduct are thestudents’ responsibility. The instructor reserves the right to make any changes to the course throughout thesemester. The only way to succeed in this course is to do the work.

1

Objectives of the Course

The objectives of the course are to learn analysis and learn proofs.

Questions: What is analysis? How is it different from other branches of mathematics?

Analysis is the branch of mathematics that studies limit and concepts derived from limit, such as continuity,differentiation and integration, while geometry deals with figures, algebra deals with equations and inequalitiesinvolving addition, subtraction, multiplication, division and number theory studies integers.

When we try to solve problems involving real or complex numbers, such as finding roots of polynomials orsolving differential equations, we may not get the right answers the first time. However, we can get approximationsand the limits of these approximations, we hope, will give us the right answers. At least, we can know solutions existeven though we may not be able to write them explicitly.

Problems involving integers can be much harder since integers are discrete, i.e. there are minimum distanceseparating distinct integers so that one cannot find integers arbitrarily close to another integer.

Try to see if there is a real solution to the equation2x � x2

4x4 C 1D 987654321: Then try to see if there is an integer

solution. What are the differences in the way you solve these two problems?

Question: Why should we learn proofs?

A statement is true not because your teacher tells you it is true. A teacher can make mistakes! There were famousmathematicians who made conjectures that were discovered to be wrong years later. How can we be certain the factswe learned are true? How can we judge when more than one proposed solutions are given, which is correct?

Suppose we want to find limx!0

x2 sin 1x

sin x: Since the numerator is between �x2 and x2; the numerator has limit 0.

The denominator also has limit 0. So, by l’Hopital’s rule, limx!0

x2 sin 1x

sin xD lim

x!0

2x sin 1x � cos 1

x

cos x: However, the new

numerator does not have a limit because cos 1x has no limit as x ! 0; while the new denominator has limit 1. So the

limit of the original problem does not exist. Is this reasoning correct? No. Where is the mistake?

Sometimes we explain facts by examples or pictures. For instance, the statement that every odd degree polynomialwith real coefficients must have at least one root is often explained by some examples or some pictures. In our lifetime,we can only do finitely many examples and draw finitely many pictures. Should we believe something is true by seeinga few pictures or examples?

Draw the graphs of a few continuous functions on [0; 1]: Do you think every continuous function on [0; 1]is differentiable in at least one point on .0; 1/? Or do you think there exists a continuous function on [0; 1] notdifferentiable at any point of .0; 1/?

Consider the function f .n/ D n2 � n C 41: Note f .1/ D 41; f .2/ D 43; f .3/ D 47; f .4/ D 53; f .5/ D61; f .6/ D 71; f .7/ D 83; f .8/ D 97; f .9/ D 113; f .10/ D 131 are prime numbers. Should you believe that f .n/is a prime number for every positive integer n? What is the first n that f .n/ is not prime?

In order to have confident, you have to be able to judge the facts you learned are absolutely correct. Almostcorrect is not good enough in mathematics.

2

Chapter 1. Logic

To reason correctly, we have to follow some rules. These rules of reasoning are what we called logic. We willonly need a few of these rules, mainly to deal with taking opposite of statements and to handle conditional statements.

We will use the symbol � (or :) to denote the word “not”. Also, we will use the symbol 8 to denote “for all”,“for any”, “for every”. Similarly, the symbol 9 will denote “there is (at least one)”, “there exists”, “there are (some)”and usually followed by “such that”. The symbols 8 and 9 are called quantifiers.

Negation. Below we will look at rules of negation (i.e. taking opposite). They are needed when we do indirectproofs (or proofs by contradiction). For any expression p; we have � .� p/ D p:Examples. (1)

expression :

pz }| {x > 0 and

qz }| {x < 1

opposite expression : x � 0 or x � 1

rule : � .p and q/ D .� p/ or .� q/(2) expression : x < 0 or x > 1

opposite expression : x � 0 and x � 1

rule : � .p or q/ D .� p/ and .� q/(3) statement : For every x � 0; x has a square root. (True)

quantified statement : 8x � 0 .x has a square root/:opposite statement : There exists x � 0 such that x does not have a square root. (False)

quantified opposite statement : 9x � 0 � .x has a square root/:(4) statement : For every x � 0; there is y � 0 such that y2 D x : (True)

quantified statement : 8x � 0; 9y � 0 .y2 D x/:opposite statement : There exists x � 0 such that for every y � 0; y2 6D x : (False)

quantified opposite statement : 9x � 0; 8y � 0 � .y2 D x/:From examples (3) and (4), we see that the rule for negating statements with quantifiers is first switch every8 to 9 and every 9 to 8, then negate the remaining part of the statement.

If-then Statements. If-then statements occur frequently in mathematics. We will need to know some equivalentways of expressing an if-then statement to do proofs. The statement “if p; then q” may also be stated as “p implies q”,“p only if q”, “p is sufficient for q”, “q is necessary for p” and is commonly denoted by “p H) q”. For example,the statement “if x D 3 and y D 4; then x2 C y2 D 25” may also be stated as “x D 3 and y D 4 are sufficient forx2 C y2 D 25” or “x2 C y2 D 25 is necessary for x D 3 and y D 4”.

Example. (5) statement : If x � 0; then jxj D x : (True)

opposite statement : x � 0 and jxj 6D x : (False)

rule : � .p H) q/ D p and .� q/Remark. Note

p H) q D� .� .p H) q//D� .p and .� q//D .� p/ or � .� q/D .� p/ or q:3

For the statement “if p; then q” .p H) q/; there are two related statements: the converse of the statement is “ifq , then p” (q H) p) and the contrapositive of the statement is “if .� q/; then .� p/” (� q H) � p).

Examples. (6) statement : If x D �3; then x2 D 9: (True)

converse : If x2 D 9; then x D �3: (False, as x may be 3.)

contrapositive : If x2 6D 9; then x 6D �3: (True)

(7) statement : x D �3 H) 2x D �6 (True)

converse : 2x D �6 H) x D �3 (True)

contrapositive : 2x 6D �6 H) x 6D �3 (True)

(8) statement : If jxj D 3; then x D �3: (False, as x may be 3.)

converse : If x D �3; then jxj D 3: (True)

contrapositive : If x 6D �3; then jxj 6D 3: (False, as x may be 3.)

Remarks. Examples (6) and (7) showed that the converse of an if-then statement is not the same as the statementnor the opposite of the statement in general. Examples (6), (7) and (8) showed that an if-then statement and itscontrapositive are either both true or both false. In fact, this is always the case because by the remark on the last page,.� q/ H) .� p/ D � .� q/ or .� p/D q or .� p/D .� p/ or qD p H) q:So an if-then statement and its contrapositive statement are equivalent.

Finally, we introduce the terminology “p if and only if q” to mean “if p; then q” and “if q; then p”. The statement“p if and only if q” is the same as “p is necessary and sufficient for q”. We abbreviate “p if and only if q” by “p ()q”. So p () q means p H) q and q H) p”. The phrase “if and only if” is often abbreviated as “iff”.

Caution! Note 8�8� D 8�8� and 9�9� D 9�9�; but 8�9� 6D 9�8�: For example, “every student is assigned anumber” is the same as “8 student, 9 number such that the student is assigned the number.” This statement impliesdifferent students may be assigned possibly different numbers. However, if we switch the order of the quantifiers, thestatement becomes “9 number such that 8 student, the student is assigned the number.” This statement implies thereis a number and every student is assigned that same number!

4

Chapter 2. Sets

To read and write mathematical expressions accurately and concisely, we will introduce the language of sets. Aset is a collection of “objects” (usually numbers, ordered pairs, functions, etc.) If object x is in a set S, then we say x isan element (or a member) of S and write x 2 S: If x is not an element of S, then we write x 62 S: A set having finitelymany elements is called a finite set, otherwise it is called an infinite set. The empty set is the set having no objects andis denoted by ;:

A set may be shown by listing its elements enclosed in braces (eg. f1; 2; 3g is a set containing the objects 1; 2; 3; thepositive integer N D f1; 2; 3; : : :g, the integerZD f: : : ;�2;�1; 0; 1; 2; : : :g, the empty set ; D fg) or by descriptionenclosed in braces (eg. the rational numbersQD fm

n : m 2Z; n 2 Ng; the real numbersRD fx : x is a real numbergand the complex numbers C D fx C iy : x; y 2 Rg:) In describing sets, the usual convention is to put the form of theobjects on the left side of the colon and to state the conditions on the objects on the right side of the colon. The setwill consist of all elements satisifying all the conditions. It is also common to use a vertical bar in place of colon in setdescriptions.

Examples. (i) The closed interval with endpoints a; b is [a; b] D fx : x 2 R and a � x � bg:(ii) The set of square numbers is f1; 4; 9; 16; 25; : : :g D fn2 : n 2 Ng:

(iii) The set of all positive real numbers isRC D fx : x 2 R and x > 0g: (If we want to emphasize this is a subsetofR; we may stress x is real in the form of the objects and writeRC D fx 2 R : x > 0g: If numbers are alwaystaken to mean real numbers, then we may write simplyRC D fx : x > 0g:)

(iv) The set of points (or ordered pairs) on the line `m with equation y D mx is f.x;mx/ : x 2 Rg:For sets A; B; we say A is a subset of B (or B contains A) iff every element of A is also an element of B: In

that case, we write A � B: (For the case of the empty set, we have ; � S for every set S:) Two sets A and B areequal if and only if they have the same elements (i.e. A D B means A � B and B � A:) So A D B if and only if(x 2 A () x 2 B). If A � B and A 6D B; then we say A is a proper subset of B and write A � B: (For example,if A D f1; 2g; B D f1; 2; 3g;C D f1; 1; 2; 3g; then A � B is true, but B � C is false. In fact, B D C: Repeatedelements are counted only one time so that C has 3 elements, not 4 elements.)

For a set S; we can collect all its subsets. This is called the power set of S and is denoted by P.S/ or 2S : Forexamples, P.;/ D f;g; P.f0g/ D f;; f0gg and P.f0; 1g/ D f;; f0g; f1g; f0; 1gg: For a set with n elements, its powerset will have 2n element. This is the reason for the alternative notation 2S for the power set of S: Power set is oneoperation of a set. There are a few other common operations of sets.

Definitions. For sets A1; A2; : : : ; An ;(i) their union is A1 [ A2 [ � � � [ An D fx : x 2 A1 or x 2 A2 or � � � or x 2 Ang;

(ii) their intersection is A1 \ A2 \ � � � \ An D fx : x 2 A1 and x 2 A2 and � � � and x 2 Ang;(iii) their Cartesian product is

A1 � A2 � � � � � An D f.x1; x2; : : : ; xn/ : x1 2 A1 and x2 2 A2 and � � � and xn 2 Ang;(iv) the complement of A2 in A1 is A1 n A2 D fx : x 2 A1 and x 62 A2g:Examples. (i) f1; 2; 3g [ f3; 4g D f1; 2; 3; 4g; f1; 2; 3g \ f2; 3; 4g D f2; 3g; f1; 2; 3g n f2; 3; 4g D f1g:(ii) [�2; 4]\N D f1; 2; 3; 4g; [0; 2][ [1; 5][ [4; 6] D [0; 6]:

(iii) .[0; 7] \Z/ n fn2 : n 2 Ng D f0; 1; 2; 3; 4; 5; 6;7g n f1; 4; 9; 16; 25; : : :g D f0; 2; 3; 5; 6; 7g:(iv) R�R�RD f.x; y; z/ : x; y; z 2 Rg;Q� .RnQ/ D f.a; b/ : a is rational and b is irrationalg:Remarks. (i) For the case of the empty set, we have

A [ ; D A D ; [ A; A \ ; D ; D ; \ A; A �; D ; D ; � A; A n ; D A and ; n A D ;:5

(ii) The notions of union, intersection and Cartesian product may be extended to infinitely many sets similarly. Theunion is the set of objects in at least one of the sets. The intersection is the set of objects in every one of the sets.The Cartesian product is the set of ordered tuples such that the i-th coordinate must belong to the i-th set.

(iii) The set A1 [ A2 [ � � � [ An may be written asn[

kD1

Ak : If for every positive integer k; there is a set Ak ; then the

notation A1 [ A2 [ A3 [ � � � may be abbreviated as1[

kD1

Ak or[k2N Ak : If for every x 2 S; there is a set Ax ; then

the union of all the sets Ax ’s for all x 2 S is denoted by[x2S

Ax : Similar abbreviations exist for intersection and

Cartesian product.

Examples. (i) .[1; 2][ [2; 3][ [3; 4][ [4; 5][ � � �/ \ZD [1;C1/ \ZD N:(ii)

\n2Nh0; 1C 1

n

� D [0; 2/ \ h0; 11

2

� \ h0; 11

3

� \ h0; 11

4

�\ � � � D [0; 1]:(iii) For every k 2 N; let Ak D f0; 1g; then

A1 � A2 � A3 � � � � D f.x1; x2; x3; : : :/ : each xk is 0 or 1 for k D 1; 2; 3; : : :g:(iv) For each m 2 R; let `m be the line with equation y D mx on the plane, then

[m2R`m D R2 n f.0; y/ : y 2 R; y 6D 0g

and\

m2R`m D f.0; 0/g:(v) Show that if A � B and C � D; then A \ C � B \ D:

Reason. For every x 2 A \C; we have x 2 A and x 2 C: Since A � B and C � D; we have x 2 B and x 2 D;which imply x 2 B \ D: Thus, we see that every element in A \C is also in B \ D: Therefore, A \C � B \ D:

(vi) Show that .A [ B/ n C D .A n C/ [ .B n C/:Reason. For every x 2 .A[ B/nC; we have x 2 A[ B and x 62 C: So either x 2 A or x 2 B: In the former case,x 2 A n C or in the latter case, x 2 B n C: So x 2 .A n C/ [ .B n C/: Hence, .A [ B/ n C � .A nC/ [ .B n C/:

Conversely, for every x 2 .A n C/ [ .B n C/; either x 2 A n C or x 2 B n C: In the former case, x 2 Aand x 62 C or in the latter case, x 2 B and x 62 C: In both cases, x 2 A [ B and x 62 C: So x 2 .A [ B/ n C:Hence, .A n C/ [ .B n C/ � .A [ B/ n C: Combining with the conclusion of the last paragraph, we have.A [ B/ n C D .A n C/ [ .B n C/:We shall say that sets are disjoint iff their intersection is the empty set. Also, we say they are mutually disjoint iff

the intersection of every pair of them is the empty set. A relation on a set E is any subset of E � E : The following isan important concept that is needed in almost all branches of mathematics. It is a tool to divide (or partition) the set ofobjects we like to study into mutually disjoint subsets.

Definition. An equivalence relation R on a set E is a subset R of E � E such that

(a) (reflexive property) for every x 2 E; .x; x/ 2 R;(b) (symmetric property) if .x; y/ 2 R; then .y; x/ 2 R;(c) (transitive property) if .x; y/; .y; z/ 2 R; then .x; z/ 2 R:

We write x � y if .x; y/ 2 R: For each x 2 E; let [x] D fy : x � yg: This is called the equivalence classcontaining x : Note that every x 2 [x] by (a) so that

[x2E

[x] D E : If x � y; then [x] D [y] because by (b) and (c),

z 2 [x] () z � x () z � y () z 2 [y]: If x 6� y; then [x] \ [y] D ; because assuming z 2 [x] \ [y] willlead to x � z and z � y; which imply x � y; a contradiction. So every pair of equivalence classes are either the sameor disjoint. Therefore, R partitions the set E into mutually disjoint equivalence classes.

6

Examples. (1) (Geometry) For triangles T1 and T2; define T1 � T2 if and only if T1 is similar to T2: This is anequivalence relation on the set of all triangles as the three properties above are satisfied. For a triangle T; [T ] is the setof all triangles similar to T :(2) (Arithmetic) For integers m and n; define m � n if and only if m � n is even. Again, properties (a), (b), (c) caneasily be verified. So this is also an equivalence relation on Z: There are exactly two equivalence classes, namely[0] D f: : : ;�4;�2; 0; 2; 4; : : :g (even integers) and [1] D f: : : ;�5;�3;�1; 1; 3; 5; : : :g (odd integers). Two integersin the same equivalence class is said to be of the same parity.

(3) Some people think that properties (b) and (c) imply property (a) by using (b), then letting z D x in (c) to conclude.x; x/ 2 R: This is false as shown by the counterexample that E D f0; 1g and R D f.1; 1/g; which satisfies properties(b) and (c), but not property (a). R fails property (a) because 0 2 E; but .0; 0/ 62 R as 0 is not in any ordered pair in R:

A function (or map or mapping) f from a set A to a set B (denoted by f : A ! B) is a method of assigning toevery a 2 A exactly one b 2 B: This b is denoted by f .a/ and is called the value of f at a: Thus, a function must bewell-defined in the sense that if a D a0; then f .a/ D f .a0/: The set A is called the domain of f (denoted by dom f )and the set B is called the codomain of f (denoted by codom f ). We say f is a B-valued function (eg. if B D R; thenwe say f is a real-valued function.) When the codomain B is not emphasized, then we may simply say f is a functionon A: The image or range of f (denoted by f .A/ or im f or ran f ) is the set f f .x/ : x 2 Ag: (To emphasize this is asubset of B; we also write it as f f .x/ 2 B : x 2 Ag:) The set G D f�x; f .x/� : x 2 Ag is called the graph of f: Twofunctions are equal if and only if they have the same graphs. In particular, the domains of equal functions are the sameset.

Examples. The function f : Z! R given by f .x/ D x2 has dom f D Z; codom f D R: Also, ran f Df0; 1; 4; 9; 16; : : :g: This is different from the function g : R! Rgiven by g.x/ D x2 because dom g D R 6D dom f:Also, a function may have more than one parts in its definition, eg. the absolute value function h : R! Rdefined by

h.x/ D nx if x � 0�x if x < 0

: Be careful in defining functions. The following is bad: let xn D .�1/n and i.xn/ D n: The

rule is not well-defined because x1 D �1 D x3; but i.x1/ D 1 6D 3 D i.x3/:Definitions. (i) The identity function on a set S is IS : S ! S given by IS.x/ D x for all x 2 S:(ii) Let f : A ! B; g : B 0 ! C be functions and f .A/ � B 0: The composition of g by f is the function

g � f : A ! C defined by .g � f /.x/ D g. f .x// for all x 2 A:(iii) Let f : A ! B be a function and C � A: The function f jC : C ! B defined by f jC.x/ D f .x/ for every x 2 C

is called the restriction of f to C:(iv) A function f : A ! B is surjective (or onto) iff f .A/ D B:(v) A function f : A ! B is injective (or one-to-one) iff f .x/ D f .y/ implies x D y:

(vi) A function f : A ! B is a bijection (or a one-to-one correspondence) iff it is injective and surjective.

(vii) For an injective function f : A ! B; the inverse function of f is the function f �1 : f .A/ ! A defined byf �1.y/ D x () f .x/ D y:

Remarks. A function f : A ! B is surjective means f .A/ D B; which is the same as saying every b 2 B is an f .a/for at least one a 2 A: In this sense, the values of f do not omit anything in B: We will loosely say f does not omitany element of B for convenience. However, there may possibly be more than one a 2 A that are assigned the sameb 2 B: Hence, the range of f may repeat some elements of B: If A and B are finite sets, then f surjective implies thenumber of elements in A is greater than or equal to the number of elements in B:

Next, a function f : A ! B is injective means, in the contrapositive sense, that x 6D y implies f .x/ 6D f .y/;which we may loosely say f does not repeat any element of B: However, f may omit elements of B as there maypossibly be elements in B that are not in the range of f: So if A and B are finite sets, then f injective implies thenumber of elements in A is less than or equal to the number of elements in B:

Therefore, a bijection from A to B is a function whose values do not omit nor repeat any element of B: If A andB are finite sets, then f bijective implies the number of elements in A and B are the same.

7

Remarks (Exercises). (a) Let f : A ! B be a function. We have f is a bijection if and only if there is a functiong : B ! A such that g � f D IA and f � g D IB : (In fact, for f bijective, we have g D f �1 is bijective.)

(b) If f : A ! B and h : B ! C are bijections, then h � f : A ! C is a bijection.

(c) Let A; B be subsets ofRand f : A ! B be a function. If for every b 2 B; the horizontal line y D b intersects thegraph of f exactly once, then f is a bijection.

Example. Show that f : [0; 1] ! [3; 4] defined by f .x/ D x3 C 3 is a bijection.

Method 1 If f .x/ D f .y/; then x3 C 3 D y3 C 3; which implies x3 D y3: Taking cube roots of both sides, we getx D y: Hence f is injective. Next, for every y 2 [3; 4]; solving the equation x3 C 3 D y for x; we get x D 3

py � 3:

Since y 2 [3; 4] implies y � 3 2 [0; 1]; we see x 2 [0; 1]: Then f .x/ D y: Hence f is surjective. Therefore, f isbijective.

Method 2. Define g : [3; 4] ! [1; 2] by g.y/ D 3p

y � 3: For every x 2 [0; 1] and y 2 [3; 4]; .g� f /.x/ D g.x3C3/ D3p.x3 C 3/� 3 D x and . f � g/.y/ D f . 3

py � 3/ D . 3

py � 3/3 C 3 D y: By remark (a) above, f is a bijection.

To deal with the number of elements in a set, we introduce the following concept. For sets S1 and S2; we willdefine S1 � S2 and say they have the same cardinality (or the same cardinal number) if and only if there exists abijection from S1 to S2: This is easily checked to be an equivalence relation on the collection of all sets. For a setS; the equivalence class [S] is often called the cardinal number of S and is denoted by card S or jSj: This is a wayto assign a symbol for the number of elements in a set. It is common to denote card ; D 0; , for a positive integern; card f1; 2; : : : ; ng D n; card N D @0 (read aleph-naught) and card R D c (often called the cardinality of thecontinuum).

8

Chapter 3. Countability

Often we compare two sets to see if they are different. In case both are infinite sets, then the concept of countablesets may help to distinguish these infinite sets.

Definitions. A set S is countably infinite iff there exists a bijection f : N! S (i.e. N and S have the same cardinalnumber @0:) A set is countable iff it is a finite or countably infinite set. A set is uncountable iff it is not countable.

Remarks. Suppose f :N! S is a bijection. Then f is injective means f .1/; f .2/; f .3/; : : : are all distinct and fis surjective means f f .1/; f .2/; f .3/; : : :g D S: So n 2 N$ f .n/ 2 S is a one-to-one correspondence between Nand S: Therefore, the elements of S can be listed in an “orderly” way (as f .1/; f .2/; f .3/; : : :) without repetition oromission. Conversely, if the elements of S can be listed as s1; s2; : : : without repetition or omission, then f : N! Sdefined by f .n/ D sn will be a bijection as no repetition implies injectivity and no omission implies surjectivity.

Bijection Theorem. Let g: S ! T be a bijection. S is countable if and only if T is countable.

(Reasons. The finite set case is clear. For infinite sets, it is true because S countable implies there is a bijective functionf : N! S; which implies h D g � f : N! T is bijective, i.e. T is countable. For the converse, h is bijective impliesf D g�1 � h is bijective.)

Remarks. Similarly, taking contrapositive, S is uncountable if and only if T is uncountable.

Basic Examples. (1) N is countably infinite (because the identity function IN.n/ D n is a bijection).

(2) Zis countably infinite because the following function is a bijection (one-to-one correspondence):N D f 1; 2; 3; 4; 5; 6; 7; 8; 9; : : : gf # # # # # # # # # # : : :Z D f 0; 1; �1; 2; �2; 3; �3; 4; �4; : : : g:

The function f : N!Zis given by f .n/ D � n2 if n is even�. n�1

2 / if n is oddand its inverse function g : Z! N is given

by g.m/ D �2m if m > 01� 2m if m � 0

: Just check g � f D IN and f � g D IZ:(3) N�N D f.m; n/: m; n 2 Ng is countably infinite.

(Diagonal Counting Scheme) Using the diagram on the right, de-fine f :N! N� N by f .1/ D .1; 1/, f .2/ D .2; 1/, f .3/ D .1; 2/,f .4/ D .3; 1/, f .5/ D .2; 2/, f .6/ D .1; 3/, : : :, then f is injectivebecause no ordered pair is repeated. Also, f is surjective because.m; n/ D f

mCn�2X

kD0

k C n

! D f

� .m C n � 2/.m C n � 1/2

C n

� : .1; 1/ .1; 2/ .1; 3/ .1; 4/.2; 1/ .2; 2/ .2; 3/ ::::::.3; 1/ .3; 2/ ::: :::::: :::.4; 1/ ::: ::: :::::: ::: ::: :::(4) The open interval .0; 1/ D fx : x 2 R and 0 < x < 1g is uncountable. Also,R is uncountable.

f .1/ D 0:a11a12a13a14 : : :f .2/ D 0:a21a22a23a24 : : :f .3/ D 0:a31a32a33a34 : : :f .4/ D 0:a41a42a43a44 : : :::: ::: Suppose .0; 1/ is countably infinite and f :N ! .0; 1/ is a bijection as

shown on the left. Consider the number x whose decimal representation is

0:b1b2b3b4 : : :, where bn D �2 if ann D 11 if ann 6D 1

. Then 0 < x < 1 and x 6D f .n/for all n because bn 6D ann . So f cannot be surjective, a contradiction. NextRis uncountable because tan�.x � 1

2 / provides a bijection from .0; 1/ ontoR:To determine the countability of more complicated sets, we will need the theorems below.

9

Countable Subset Theorem. Let A � B: If B is countable, then A is countable. (Taking contrapositive, if A isuncountable, then B is uncountable.)

Countable Union Theorem. If An is countable for every n 2 N; then[n2N An is countable. In general, if S is countable

(say f : N! S is a bijection) and As is countable for every s 2 S; then[s2S

As D [n2N A f .n/ is countable. (Briefly,

countable union of countable sets is countable.)

Product Theorem. If A, B are countable, then A�B D f.a; b/: a 2 A; b 2 Bg is countable. In fact, if A1; A2; : : : ; An

are countable, then A1 � A2 � � � � � An is countable (by mathematical induction).

(Sketch of Reasons. For the countable subset theorem, if B is countable, then we can list the elements of B and tocount the elements of A, we can skip over those elements of B that are not in A: For the countable union theorem, ifwe list the elements of A1 in the first row, the elements of A2 in the second row, : : : ; then we can count all the elementsby using the diagonal counting scheme. As for the product theorem, we can imitate the example of N� N and alsouse the diagonal counting scheme.)

Examples. (5)QD 1[nD1

Sn;where Sn D nm

n: m 2Zo:For every n 2 N; the function fn : Z! Sn given by fn.m/ D m

nis a bijection (with f �1

n

�mn

� D m), so Sn is countable by the bijection theorem. Therefore, Q is countable by the

countable union theorem. (Then subsets ofQ likeZn f0g;N[ f0g;Q\ .0; 1/ are also countable.)

(6) R n Q is uncountable. (In fact, if A is uncountable and B is countable, then A n B is uncountable as A n Bcountable implies .A \ B/ [ .A n B/ D A countable by the countable union theorem, which is a contradiction).

(7) C D fx C iy : x; y 2 Rg containsRandR is uncountable, so by the countable subset theorem, C is uncountable.

(8) Show that the set A D frpm : m 2 N; r 2 .0; 1/g is uncountable, but the set B D frpm : m 2 N; r 2 Q\ .0; 1/gis countable.

Solution. Taking m D 1; we see that .0; 1/ � A: Since .0; 1/ is uncountable, A is uncountable. Next we willobserve that B D [

m2N Bm; where Bm D frpm : r 2 Q \ .0; 1/g D [r2Q\.0;1/frpmg for each m 2 N: SinceQ\ .0; 1/ is countable and frpmg has 1 element for every r 2 Q\ .0; 1/; Bm is countable by the countable union

theorem. Finally, since N is countable and Bm is countable for every m 2 N; B is countable by the countableunion theorem.

(9) Show that the set L of all lines with equation y D mx C b; where m; b 2 Q; is countable.

Solution. Note that for each pair m; b of rational numbers, there is a unique line y D mx C b in the set L : So thefunction f : Q�Q! L defined by letting f .m; b/ be the line y D mx C b (with f �1 sending the line back to.m; b/) is a bijection. SinceQ�Q is countable by the product theorem, so the set L is countable by the bijectiontheorem.

(10) Show that if An D f0; 1g for every n 2 N; then A1 � A2 � A3 � � � � is uncountable. (In particular, this shows thatthe product theorem is not true for infintely many countable sets.)

Solution. Assume A1 � A2 � A3 � � � � D f.a1; a2; a3; : : :/ : each ai D 0 or 1g is countable and f : N !A1 � A2 � A3 � � � � is a bijection. Following example (4), we can change the n-th coordinate of f .n/ (from 0 to1 or from 1 to 0) to produce an element of A1 � A2 � A3 � � � � not equal to any f .n/; which is a contradiction.So it must be uncountable.

(11) Show that the power set P.N/ of all subsets of N is uncountable.

Solution. As in example (10), let An D f0; 1g for every n 2 N: Define g : P.N/ ! A1 � A2 � A3 � � � � by

g.S/ D .a1; a2; a3; : : :/; where am D �1 if m 2 S0 if m 62 S

: (For example, g.f1; 3; 5; : : :g/ D .1; 0; 1; 0; 1; : : :/:) Note

g has the inverse function g�1�.a1; a2; a3; : : :/� D fm : am D 1g: Hence g is a bijection. Since A1� A2� A3�� � �

is uncountable, so P.N/ is uncountable by the bijection theorem.

10

(12) Show that the set S of all nonconstant polynomials with integer coefficients is countable.Solution. For n 2 N; the set of Sn of all polynomials of degree n with integer coefficients is countable becausethe function f : Sn ! .Zn f0g/ �Z� � � � �Zdefined by f .anxn C an�1xn�1 C � � � C a0/ D .an; an�1; : : : ; a0/is a bijection and .Zn f0g/ �Z� � � � �Zis countable by the product theorem. So, S D [

n2N Sn is countable by

the countable union theorem.

(13) Show that there exists a real number, which is not a root of any nonconstant polynomial with integer coefficients.

Solution. For every nonconstant polynomial f with integer coefficients, let R f denotes the set of roots of f: ThenR f has at most (deg f ) elements, hence R f is countable. Let S be the set of all nonconstant polynomials withinteger coefficients, which is countable by the last example. Then

[f 2S

R f is the set of all roots of nonconstant

polynomials with integer coefficients. It is countable by the countable union theorem. Since R is uncountable,Rn[f 2S

R f is uncountable by the fact in example (6). So there exist uncountably many real numbers, which are

not roots of any nonconstant polynomial with integer coefficients.

Remarks. Any number which is a root of a nonconstant polynomial with integer coefficients is called an algebraicnumber. A number which is not a root of any nonconstant polynomial with integer coefficients is called a transcendentalnumber. transcendental numbers? If so, are there finitely many or countably many such numbers? Since every rationalnumber a

b is the root of the polynomial bx � a; every rational number is algebraic. There are irrational numbers like�p2; which are algebraic because they are the roots of x2 � 2: Using the identity cos 3� D 4 cos3 � � 3 cos �; theirrational number cos 20� is easily seen to be algebraic as it is a root of 8x3 � 6x � 1: Example (13) showed that thereare only countably many algebraic numbers and there are uncountably many transcendental real numbers. In a numbertheory course, it will be shown that � and e are transcendental.

Theorem.

(1) (Injection Theorem) Let f : A ! B be injective. If B is countable, then A is countable. (Taking contrapositive,if A is uncountable, then B is uncountable.)

(2) (Surjection Theorem) Let g : A ! B be surjective. If A is countable, then B is countable. (Taking contrapositive,if B is uncountable, then A is uncountable.)

(Reasons. For the first statement, observe that the function h : A ! f .A/ defined by h.x/ D f .x/ is injective(because f is injective) and surjective (because h.A/ D f .A/). So h is a bijection. If B is countable, then f .A/ iscountable by the countable subset theorem, which implies A is countable by the bijection theorem.

For the second statement, observe that B D g.A/ D[x2A

fg.x/g: If A is countable, then it is a countable union of

countable sets. By the countable union theorem, B is countable.)

Examples. (14) ShowQ is countable by using the injection theorem.

Solution. Define f : Q!Z� N by f .x/ D .m; n/; where m=n is the reduced fraction form of x : Then f isinjective because f .x/ D f .x 0/ D .m; n/ implies x D m=n D x 0: SinceZ�N is countable by product theorem,soQ is countable by the injection theorem.

(15) Let A1 be uncountable and A2; : : : ; An be nonempty sets. Show that A1 � A2 � � � � � An is uncountable.

Solution. Define g : A1 � A2 � � � � � An ! A1 by g.x1; x2; : : : ; xn/ D x1: Since A2; : : : ; An are nonempty, leta2 2 A2; : : : ; an 2 An : Then for every a1 2 A1; we have g.a1; a2; : : : ; an/ D a1 so that g is surjective. Since A1

is uncountable, by the surjection theorem, A1 � A2 � � � � � An is uncountable.

The following is a famous statement in mathematics.

Continuum Hypothesis. If S is uncountable, then there exists at least one injective function f : R! S; i.e. everyuncountable set has at least as many elements as the real numbers.

In 1940, Kurt Godel showed that the opposite statement would not lead to any contradiction. In 1966, Paul Cohenwon the Fields’ Medal for showing the statement also would not lead to any contradiction. So proof by contradictionmay not be applied to every statement.

11

Chapter 4. Series

Definitions. A series is the summation of a countable set of numbers in a specific order. If there are finitely manynumbers, then the series is a finite series, otherwise it is an infinite series. The numbers are called terms. The sum ofthe first n terms is called the n-th partial sum of the series.

An infinite series is of the form a1|{z}1st term

C a2|{z}2nd term

C a3|{z}3rd term

C : : : or we may write it as1X

kD1

ak:The first partial sum is S1 D a1. The second partial sum is S2 D a1Ca2. The nth partial sum is Sn D a1Ca2C: : :Can .

Series are used frequently in science and engineering to solve problems or approximate solutions. (E.g. trigono-metric or logarithm tables were computed using series in the old days.)

Examples.(1) 1C 12C 1

4C 18C 1

16C: : : D? (Sn D 1C 12C 1

4C: : :C 12n D 2� 1

2n ; 1C 12C 1

4C 18C 1

16C: : : D limn!1�2� 1

2n

� D 2.)

We say the series converges to 2, which is called the sum of the series.

(2) 1C 1C 1C 1C 1C 1C : : : D 1 (Sn D 1C 1C : : : C 1| {z }n

D n, limn!1 Sn D 1.) We say the series diverges (to1).

(3) 1� 1C 1� 1C 1� 1C 1� 1C : : :. (Sn D n1 if n is odd0 if n is even

, limn!1 Sn doesn’t exist.) We say the series diverges.

Definitions. A series1X

kD1

ak D a1Ca2Ca3C : : : converges to a number S iff limn!1.a1Ca2C : : :Can/ D lim

n!1 Sn D S:In that case, we may write

1XkD1

ak D S and say S is the sum of the series. A series diverges to1 iff the partial sum Sn

tends to infinity as n tends to infinity. A series diverges iff it does not converge to any number.

Remarks. (1) For every series1X

kD1

ak , there is a sequence (of partial sums) fSng. Conversely, if the partial sum sequencefSng is given, we can find the terms an as follows: a1 D S1, a2 D S2 � S1, : : :, ak D Sk � Sk�1 for k > 1. Then

a1C : : :C an D S1C .S2� S1/C : : :C .Sn � Sn�1/ D Sn. So fSng is the partial sum sequence of1X

kD1

ak . Conceptually,

series and sequences are equivalent. So to study series, we can use facts about sequences.

(2) Let N be a positive integer.1X

kD1

ak converges to A if and only if1X

kDN

ak converges to B D A � .a1 C � � � C aN�1/because

B D limn!1.aN C � � � C an/ D lim

n!1.a1 C a2 C � � � C an/� .a1 C � � � C aN�1/ D A� .a1 C � � � C aN�1/:So to see if a series converges, we may ignore finitely many terms.

Theorem. If1X

kD1

ak converges to A and1X

kD1

bk converges to B; then1XkD1

.ak C bk/ D AC B D 1XkD1

ak C 1XkD1

bk; 1XkD1

.ak � bk/ D A� B D 1XkD1

ak � 1XkD1

bk; 1XkD1

cak D cA D c1X

kD1

ak

for any constant c:For simple series such as geometric or telescoping series, we can find their sums.

12

Theorem (Geometric Series Test). We have1XkD0

rk D limn!1.1C r C r2 C : : : C rn/ D lim

n!1 1� rnC1

1� rD (

1

1� rif jrj < 1

doesn’t exist otherwise:

Example. 0:999 � � � D 9

10C 9

100C 9

1000C � � � D 9

10. 1

1� 110

/ D 1 D 1:000 � � � : So 1 has two decimal representa-

tions!

Theorem (Telescoping Series Test). We have1X

kD1

.bk � bkC1/ D limn!1�.b1 � b2/ C .b2 � b3/ C � � � C .bn � bnC1/�D lim

n!1.b1 � bnC1/ D b1 � limn!1 bnC1 converges if and only if lim

n!1 bn is a number.

Examples. (1)1X

kD1

1

k.k C 1/ D 1XkD1

.1

k� 1

k C 1/ D �

1� 1

2

�C �1

2� 1

3

�C �1

3� 1

4

�C � � � D 1� limn!1 1

n C 1D 1:

(2)1X

kD1

.51=k � 51=.kC1// D .5�p5/C .p5� 3

p5/ C � � � D 5� lim

k!1 51=.kC1/ D 5� 50 D 4:If a series is not geometric or telescoping, we can only determine if it converges or diverges. This can be done

most of the time by applying some standard tests. If the series converges, it may be extremely difficult to find the sum!

Theorem (Term Test). If1X

kD1

ak converges, then limk!1 ak D 0. (If lim

k!1 ak 6D 0, then the series1X

kD1

ak diverges.) If

limk!1 ak D 0, the series

1XkD1

ak may or may not converge.

(Reason. Suppose1X

kD1

ak converges to S: Then limn!1 Sn D S and lim

k!1 ak D limk!1.Sk � Sk�1/ D S � S D 0.)

Term test is only good for series that are suspected to be divergent!

Examples. (1) 1C 1C 1C 1C : : :. Here ak D 1 for all k, so limk!1 ak D 1. Series diverges.

(2)1X

kD1

cos.1k/ D cos 1C cos

12C cos

13C : : : diverges because lim

k!1 cos.1k/ D cos 0 D 1 6D 0.

(3)1X

kD1

cos k D cos 1 C cos 2 C cos 3 C : : : diverges because limk!1 cos k 6D 0: (Otherwise, lim

k!1 cos k D 0: Then

limk!1 j sin kj D lim

k!1p1� cos2 k D 1 and 0 D limk!1 j cos.k C 1/j D lim

k!1 j cos k cos 1� sin k sin 1j D sin 1 6D 0; a

contradiction.)

(4) 1� 12 C 1

4 � 18 C : : :. Here ak D .� 1

2 /k�1 for all k, so limk!1 ak D 0: (Term test doesn’t apply!) Series converges

by the geometric series test.

(5) 1 C 12C 1

2| {z }2 times

C 14C 1

4C 1

4C 1

4| {z }4 times

C 18C : : : C 1

8| {z }8 times

C : : :. We have limk!1 ak D 0: (Term test doesn’t apply.) Series

diverges to1 because S1 � S2 � S3 � � � � and S2n�1 D n has limit1:13

For a nonnegative series1X

kD1

ak (i.e. ak � 0 for every k), we have S1 � S2 � S3 � : : : and limn!1 Sn must exist as a

number or equal toC1. So either1X

kD1

ak converges to a number or1X

kD1

ak diverges toC1. (In short, either1X

kD1

ak D S

or1X

kD1

ak D C1.) For nonnegative series, we have the following tests.

Theorem (Integral Test). Let f : [1;C1/! Rdecrease to 0 as x !C1. Then1X

kD1

f .k/ converges if and only ifZ 11

f .x/ dx <1. (Note in general,1X

kD1

f .k/ 6D Z 11

f .x/ dx :)(Reason. This follows from f .2/ C f .3/ C : : : C f .n/ C � � � � Z 1

1f .x/dx � f .1/ C f .2/ C : : : C f .n � 1/ C � � �

as shown in the figures below.)

...

1 2 3 4 n...

(2)

(3)

1 2 3 4 ... n

...

(2)

(1)f

f

f

f

Examples. (1) Consider the convergence or divergence of1X

kD1

1

1C k2:

As x % 1; 1 C x2 % 1; so1

1C x2& 0: Now

Z 11

1

1C x2dx D arctan x

����11

D �2� �

4<1: So

1XkD1

1

1C k2

converges.

(2) Consider the convergence or divergence of1X

kD2

1k ln k

and1X

kD2

1k.ln k/2

.

As x % 1; x ln x and x.ln x/2 % 1; so their reciprocals decrease to 0: NowZ 1

2

dx

x ln xD ln.ln x/����1

2

D 1: So1XkD2

1k ln k

diverges. NextZ 1

2

dx

x.ln x/2D � 1

ln x

����12

D 1ln 2

<1: So1X

kD2

1k.ln k/2

converges.

Theorem (p-test). For a real number p; �.p/ D 1XkD1

1

k pD 1C 1

2pC 1

3pC 1

4pC : : : converges if and only if p > 1.

(Reason. For p � 0; the terms are at least 1, so the series diverges by term test. For p > 0; f .x/ D 1x p decreases

to 0 as x ! C1. SinceZ 1

1

1

x pdx D x�pC1�p C 1

����11

D 1

p � 1if p > 1; Z 1

1

1

x pdx D .ln x/j11 D 1 if p D 1 andZ 1

1

1

x pdx D x�pC1�p C 1

����11

D 1 if p < 1, the integral test gives the conclusion.)

Remarks. For even positive integer p; the value of �.p/ was computed by Euler back in 1736. He got�.2/ D �2

6; �.4/ D �4

90; : : : ; �.2n/ D .�1/nC1 .2�/2n B2n

2.2n/! ; : : : ;14

where B0 D 1 and .k C 1/Bk D � k�1XmD0

�k C 1

m

�Bm for k � 1: The values of �.3/; �.5/; : : : are unknown. Only in the

1980’s, R. Apery was able to show �.3/ was irrational.

Theorem (Comparison Test). Given vk � uk � 0 for every k. If1X

kD1

vk converges, then1X

kD1

uk converges. If1X

kD1

uk

diverges, then1X

kD1

vk diverges.

(Reason. vk � uk � 0 ) 1XkD1

vk � 1XkD1

uk � 0. If1X

kD1

vk is a number, then1X

kD1

uk is a number. If1X

kD1

uk D C1, then1XkD1

vk D C1.)

Theorem (Limit Comparison Test). Given uk , vk > 0 for every k: If limk!1 vk

ukis a positive number L ; then either (both1X

kD1

uk and1X

kD1

vk converge) or (both diverge to C1). If limk!1 vk

ukD 0; then

1XkD1

uk converges ) 1XkD1

vk converges. If

limk!1 vk

ukD 1; then

1XkD1

uk diverges ) 1XkD1

vk diverges.

(Sketch of Reason. For k large,vk

uk� L . For L > 0;P vk � P

Luk D LP

uk. If one series converges, then the

other also converges. If one diverges (toC1), so does the other. For L D 0; vk < uk eventually. For L D 1; vk > uk

eventually. So the last two statements follow from the comparison test.)

Examples. Consider the convergence or divergence of the following series:

(1)1X

kD1

1

k2cos�1

k

�(2)

1XkD2

3k

k2 � 1(3)

1XkD1

pk C 1

k2 C 5k(4)

1XkD1

sin�1

k

�.

Solutions. (1) Since 0 < 1

k2cos�1

k

� < 1

k2and

1XkD1

1

k2converges by p-test,

1XkD1

1

k2cos�1

k

�converges.

(2) Since 0 < �3

2

�k � 3k

k2 � 1for k � 2 and

1XkD2

�3

2

�kdiverges by the geometric series test,

1XkD2

3k

k2 � 1diverges.

(3) When k is large,

pk C 1

k2 C 5k� p

k

k2D 1

k3=2. We compute lim

k!1 pkC1

k2C5kpk

k2

D limk!1r k C 1

k

k2

k2 C 5kD 1: Since

1XkD1

1

k3=2

converges by p-test,1X

kD1

pk C 1

k2 C 5kconverges by the limit comparison test.

(4) When k is large,1

kis close to 0, so sin

�1

k

�is close to

1

kbecause lim�!0

sin �� D 1 (i.e. sin � � � as � ! 0).

We compute limk!1 sin. 1

k /1k

D lim�!0

sin �� D 1. Since1X

kD1

1k

diverges by p-test,1X

kD1

sin�1

k

�diverges by the limit

comparison test.

For series with alternate positive and negative terms, we have the following test.

Theorem (Alternating Series Test). If ck decreases to 0 as k !1 (i.e. c1 � c2 � c3 � : : : � 0 and limk!1 ck D 0),

then1X

kD1

.�1/kC1ck D c1 � c2 C c3 � c4 C c5 � : : : converges.

15

(Reason. Since c1 � c2 � c3 � : : : � 0, we have 0 � S2 � S4 � S6 � : : : � S5 � S3 � S1: Since limn!1 jSn � Sn�1j D

limn!1 cn D 0, the distances between the partial sums decrease to 0 and so lim

n!1 Sn must exist.)

Examples. Both1X

kD2

.�1/k

k ln kand

1XkD1

e�k cos k� converge by the alternating series test because as k %1; k ln k %1and ek %1; so 1=.k ln k/ & 0 and e�k & 0 and cos k� D .�1/k :

For series with arbitrary positive or negative term, we have the following tests.

Theorem (Absolute Convergence Test). If1X

kD1

jak j converges, then1X

kD1

ak converges.

(Reason. From�jakj � ak � jak j; we get 0 � ak C jakj � 2jakj: Since1X

kD1

2jakj converges, so by the comparison test,1XkD1

.ak C jakj/ converges. Then1X

kD1

ak D 1XkD1

.ak C jakj/ � 1XkD1

jak j converges.)

Definition. We say1X

kD1

ak converges absolutely iff1X

kD1

jak j converges. We say1X

kD1

ak converges conditionally iff1X

kD1

ak

converges, but1X

kD1

jak j diverges.

Examples. Determine if the following series converge absolutely or conditionally

(a)1X

kD1

cos k

k3(b)

1XkD1

cos k�1C k

.

Solutions. (a)1X

kD1

����cos k

k3

���� � 1XkD1

1k3

. Since1X

kD1

1k3

converges by p-test, it follows that1X

kD1

����cos k

k3

���� converges by the

comparison test. So1X

kD1

cos k

k3converges absolutely by the absolute convergence test.

(b)1X

kD1

����cos k�1C k

���� D 1XkD1

11C k

because cos k� D .�1/k .Z 1

1

dx

1C xD ln.1C x/����1

1

D 1 ) 1XkD1

11C k

diverges.

However,1

1C kdecreases to 0 as k ! C1. So by the alternating series test,

1XkD1

cos k�1C k

D 1XkD1

.�1/k 11C k

converges. Therefore1X

kD1

cos k�1C k

converges conditionally.

Theorem (Ratio Test). If ak 6D 0 for every k and limk!1 jakC1=akj exists, then

limk!1 ����akC1

ak

���� 8>>>>>>>><>>>>>>>>:< 1 ) 1XkD1

ak converges absolutelyD 1 ) 1XkD1

ak may converge

e:g: 1X

kD1

1

k2

!or diverge

e:g: 1X

kD1

1

k

!> 1 ) 1XkD1

ak diverges

:16

(Sketch of reason. Let r D limk!1 ����akC1

ak

����, then for k large,

����akC1

ak

����, ����akC2

akC1

����, : : :, ���� akCn

akCn�1

���� � r, so jakCnj � jak jrn andjakj C jakC1j C jakC2j C : : : � jakj.1 C r C r2 C r3 C : : :/ which converges if r < 1 by the geometric series test andPakCn �P�akrn diverges if r > 1 by the term test.)

Theorem (Root Test). If limk!1 k

pjakj exists, then

limk!1 k

pjak j8>>>>>>>><>>>>>>>>:< 1 ) 1XkD1

ak converges absolutelyD 1 ) 1XkD1

ak may converge

e:g: 1X

kD1

1k2

!or diverge

e:g: 1X

kD1

1k

!> 1 ) 1XkD1

ak diverges

:(Sketch of reason. Let r D lim

k!1 kpjakj, then for k large, kpjakj � r: So jak j � rk ,

P jak j �Prk .)

Examples. Consider the convergence or divergence of the following series:

(1)1X

kD1

13k � 2k

(2)1X

kD1

k!kk

.

Solutions. (1) Since limk!1 1

3kC1�2kC1

13k�2k

D limk!1 3k � 2k

3kC1 � 2kC1D lim

k!1 13 � . 2

3/k 13

1� . 23 /kC1

D 1

3< 1; by the ratio test,

1XkD1

1

3k � 2k

converges. Alternatively, since limk!1 k

r1

3k � 2kD lim

k!1 1kp3k � 2k

D limk!1 1

3 kq

1� . 23 /k

D 1

3< 1, by the root

test,1X

kD1

1

3k � 2kconverges.

(2) Since limk!1 .k C 1/!.k C 1/kC1

kk

k!D lim

k!1 1.1 C 1k /k

D 1

e< 1; by the ratio test,

1XkD1

k!

kkconverges.

Remarks. You may have observed that in example (1), the limit you got for applying the root test was the same as thelimit you got for applying the ratio test. This was not an accident!

Theorem. If ak > 0 for all k and limk!1 akC1

akD r 2 R; then lim

k!1 kpak D r: (This implies that the root test can be

applied to more series than the ratio test.)

Examples. (1) Let ak D k; then limk!1 akC1

akD lim

k!1 k C 1

kD 1: So, lim

k!1 kpk D 1:(2) Let ak D k!

kk; then lim

k!1 akC1

akD 1

eas above. So lim

k!1 kpak D limk!1 kpk!

kD 1

e; i.e. when k is large, k! � �

k

e

�k ;which is a simple version of what is called Stirling’s formula. It is useful for estimating n! when n is large. For

example, since log10100

e� 1:566; so

100

e� 101:566; then we get 100! � 10156:6; which has about 157 digits.

Theorem (Summation by Parts). Let Sj D jXkD1

ak D a1 C a2 C : : : C aj and 1bk D bkC1 � bk.k C 1/ � kD bkC1 � bk; then

nXkD1

akbk D Snbn � n�1XkD1

Sk1bk :17

(Reason. Note a1 D S1 and ak D Sk � Sk�1 for k > 1. So,

nXkD1

akbk D S1b1 C .S2 � S1/b2 C : : : C .Sn � Sn�1/bnD Snbn � S1.b2 � b1/ � : : : � Sn�1.bn � bn�1/:/Example. Show that

1XkD1

sin k

kconverges.

Let ak D sin k and bk D 1k: Using the identity sin m sin

12D 1

2

�cos.m � 1

2/� cos.m C 1

2/� ; we have

Sk D sin 1C sin 2C � � � C sin k D cos 12 � cos.k C 1

2 /2 sin 1

2

:This implies jSkj � 1

sin.1=2/ for every k: Applying summation by parts and noting that limn!1 Sn

nD 0; we get1X

kD1

sin k

kD lim

n!1 nXkD1

sin k

kD lim

n!1 Sn

n� n�1X

kD1

Sk

�1

k C 1� 1

k

�! D 1XkD1

Sk

�1

k� 1

k C 1

� :Now

1XkD1

����Sk

�1k� 1

k C 1

����� � 1sin.1=2/ 1X

kD1

�1k� 1

k C 1

� D 1sin.1=2/ by the telescoping series test. So by the

absolute convergence test,1X

kD1

sin k

kD 1X

kD1

Sk

�1

k� 1

k � 1

�converges.

Inserting Parentheses and Rearrangements of Series.

Definition. We say1X

kD1

bk is obtained from1X

kD1

ak by inserting parentheses iff there is a strictly increasing function

p : N[f0g ! N[f0g such that p.0/ D 0; b1 D a1C� � �Cap.1/; b2 D ap.1/C1C� � �Cap.2/; b3 D ap.2/C1C� � �Cap.3/; : : : :(Note bn is the sum of kn D p.n/ � p.n � 1/ terms.)

Grouping Theorem. Let1X

kD1

bk be obtained from1X

kD1

ak by inserting parentheses. If1X

kD1

ak converges to s; then1X

kD1

bk

will converge to s: Next, if limn!1 an D 0; kn is bounded and

1XkD1

bk converges to s; then1X

kD1

ak will converge to s:(Reason. Let sn D nX

kD1

ak and tn D nXkD1

bk: For the first part,1X

kD1

bk D limn!1 tn D lim

n!1 p.n/XkD1

ak D s: For the second part,

let p.n/� p.n � 1/ be bounded by M: For a positive integer j; let p.i/ � j < p.i C 1/: For r D 1; 2; : : : ; M; define

cr; j D �ap.i/Cr if p.i/ C r � j0 if p.i/ C r > j

:Then1X

kD1

ak D limj!1 sj D lim

i!1 ti C limj!1.c1; j C � � � C cM; j / D s C 0C � � � C 0 D s:)

Examples. (1) Since1X

kD1

12kD 1

2C 1

4C 1

8C 1

16C � � � converges to 1, so by the theorem,

1

2C �1

4C 1

8

� C � 1

16C 1

32C 1

64

� C � 1

128C 1

256C 1

512C 1

1024

� C � � � D 1:18

(2) .1�1/C .1�1/C � � � converges to 0, but 1�1C1�1C � � � diverges by term test. So limn!1 an D 0 is important.

Also, .1 � 1/C �12C 1

2� 1

2� 1

2

�C �13C 1

3C 1

3� 1

3� 1

3� 1

3

�C � � � converges to 0. However, the series

without parentheses diverges (as Sn2 D 1 and Sn2Cn D 0) even though the terms have limit 0. So kn bounded isimportant.

(3) Since�

1� 1

2

�C �1

3� 1

4

�C � � � D 1XjD1

�1

2 j � 1� 1

2 j

� D 1XjD1

1

2 j .2 j � 1/ converges (by the limit compari-

son test with1X

jD1

1

j 2), so by the theorem, 1� 1

2C 1

3� 1

4C � � � D 1X

kD1

.�1/kC1

kconverges to the same sum.

Definition.1X

kD1

bk is a rearrangement of1X

kD1

ak iff there is a bijection � : N! N such that bk D a�.k/:Example. Given ln 2 D 1� 1

2C 1

3� 1

4C 1

5� 1

6C : : : (which converges conditionally). Consider the rearrangement

1C 13| {z }

2C �12|{z}

1� C 15C 1

7| {z }2C �1

4|{z}1� C 1

9C 1

11| {z }2C �1

6|{z}1� C : : :. Observe that.1 � 1

2 / C . 13 � 1

4 / C . 15 � 1

6 / C . 17 � 1

8/ C : : : D ln 2C 12 � 1

4 C 16 � 1

8 : : : D 12 ln 2

1 C . 13 � 1

2 / C 15 C . 1

7 � 14/ C : : : D 3

2 ln 2:Riemann’s Rearrangement Theorem. Let ak 2 Rand

1XkD1

ak converge conditionally. For any x 2 Ror x D �1,

there is a rearrangement1X

kD1

a�.k/ of1X

kD1

ak such that1X

kD1

a�.k/ D x :(Sketch of reason. Let pk D �

ak if ak � 00 if ak < 0

and qk D �0 if ak � 0jak j if ak < 0

: Then ak D pk � qk and jakj D pk C qk :Now both

1XkD1

pk; 1XkD1

qk must diverge toC1: (If both converges, then their sum1X

kD1

jakjwill be finite, a contradiction.

If one converges and the other diverges to C1; then1X

kD1

ak D 1XkD1

pk � 1XkD1

qk will diverges to �1; a contradiction

also.) Let un; vn be sequences of real numbers having limits x and un < vn; un < vnC1; v1 > 0: Now let P1; P2; : : :be the nonnegative terms of

1XkD1

ak in the order they occur and Q1; Q2; : : : be the absolute value of the negative terms

in the order they occur. Since1X

kD1

Pk; 1XkD1

Qk differ from1X

kD1

pk; 1XkD1

qk only by zero terms, they also diverges toC1:Let m1; k1 be the smallest integers such that P1 C � � � C Pm1 > v1 and P1 C � � � C Pm1 � Q1 � � � � � Qk1 < u1:Let m2; k2 be the smallest integers such that P1 C � � � C Pm1 � Q1 � � � � � Qk1 C Pm1C1 C � � � C Pm2 > v2 andP1C� � �C Pm1 �Q1�� � ��Qk1 C Pm1C1C� � �C Pm2 �Qk1C1�� � ��Qk2 < u2 and continue this way. This is possiblesince the sums of Pk and Qk areC1: Now if sn; tn are the partial sums of this series P1C� � �CPm1�Q1�� � ��Qk1C� � �whose last terms are Pmn ; Qkn ; respectively, then jsn � vnj � Pmn and jtn � unj � Qkn by the choices of mn; kn : SincePn; Qn have limit 0, so sn; tn must have limit x : As all other partial sums are squeezed by sn and tn; the series weconstructed must have limit x :)Dirichlet’s Rearrangement Theorem. If ak 2 Rand

1XkD1

ak converges absolutely, then every rearrangement1X

kD1

a�.k/converges to the same sum as

1XkD1

ak.

19

(Reason. Define pk; qk as in the last proof. Since pk; qk � jak j; 1XkD1

pk; 1XkD1

qk converge, say to p and q; respectively.

Since a�.k/ D p�.k/�q�.k/; we may view1X

kD1

p�.k/ as a rearrangement of the nonnegative terms of1X

kD1

ak and inserting

zeros where a�.k/ < 0: For any positive integer m; the partial sum sm D mXkD1

p�.k/ � 1XkD1

pk D p: Since pk � 0; the

partial sum sm is also increasing, hence1X

kD1

p�.k/ converges. Now, for every positive integer n; nXkD1

pk � 1XkD1

p�.k/ � p:As n !1; we get

1XkD1

p�.k/ D p: Similarly,1X

kD1

q�.k/ D q: Then1X

kD1

a�.k/ D p � q D 1XkD1

ak :)Example.

1XkD1

.�12/k D �1

2C 1

22� 1

23C 1

24� 1

25C : : : converges (absolutely) to

� 12

1� .� 12 / D �1

3.�1

2C 1

22C 1

24� 1

23| {z }2 terms

C 128� 1

27C 1

26� 1

25| {z }4 terms

C 1216

� 1215

C 1214

� 1213

C 1212

� 1211

C 1210

� 129| {z }

8 terms

C : : :is a rearrangement of

1XkD1

.�1

2/k , so it also converges to �1

3.

Remarks. As a consequence of the rearrangement theorem, the sum of a nonnegative series is the same no matter howthe terms are rearranged.

Complex Series

Complex numbers S1; S2; S3; : : : with Sn D un C ivn are said to have limit limn!1 Sn D u C iv iff lim

n!1 un D u

and limn!1 vn D v: A complex series is a series where the terms are complex numbers. The definitions of convergent,

absolutely convergent and conditional convergent are the same. The remarks and the basic properties following thedefinitions of convergent and divergent series are also true for complex series.

The geometric series test, telescoping series test, term test, absolute convergence test, ratio test and root test are

also true for complex series. For zk D xk C iyk; we have1X

kD1

zk converges to z D x C iy if and only if1X

kD1

xk converges

to x and1X

kD1

yk converges to y: So complex series can be reduced to real series for study if necessary.

Examples. (1) Note limn!1 in 6D 0 (otherwise 0 D lim

n!1 jin j D limn!1 1 is a contradiction). So

1XkD1

ik diverges by term test.

(2) If jzj � 1; then��� zk

k2

��� � 1k2

and1X

kD1

1k2

converges by p-test implies1X

kD1

zk

k2converges absolutely. However, ifjzj > 1; then lim

k!1��� zkC1.k C 1/2

k2

zk

��� D limk!1 k2.k C 1/2

jzj D jzj > 1 implies1X

kD1

zk

k2diverges by the ratio test.

20

Chapter 5. Real Numbers

Decimal representations and points on a line are possible ways of introducing real numbers, but they are not tooconvenient for proving many theorems. Instead we will introduce real numbers by its important properties.

Axiomatic Formulation. There exists a setR (called real numbers) satisfying the following four axioms:

(1) (Field Axiom) R is a field (i.e. Rhas two operationsC and � such that for any a, b, c 2 R,

(i) aCb, a �b 2 R, (ii) aCb D bCa, a �b D b �a, (iii) .aCb/Cc D aC .bCc/, .a �b/ �c D a � .b �c/,(iv) there are unique elements 0, 1 2 Rwith 1 6D 0 such that a C 0 D a, a � 1 D a,

(v) there is a unique element �a 2 Rsuch that aC .�a/ D 0; if a 6D 0; then there is a unique element a�1 suchthat a � .a�1/ D 1:

(vi) a � .b C c/ D a � b C a � c.)

(This axiom allows us to do algebra with equations. Define a � b to mean a C .�b/I ab to mean a � bI ab to

mean a � .b�1/: Also, define 2 D 1C 1; 3 D 2C 1; : : : :)(2) (Order Axiom) Rhas an (ordering) relation < such that for any a, b 2 R

(i) exactly one of the following a < b, a D b, b < a is true,

(ii) if a < b, b < c, then a < c,

(iii) if a < b, then a C c < b C c,

(iv) if a < b and 0 < c, then ac < bc.

(This axiom allows us to work with inequalities. For example, using (ii) and (iii), we can see that if a < band c < d; then aC c < bC d because aC c < bC c < bC d: Also, we can get 0 < 1 (for otherwise 1 < 0would imply by (iii) that 0 D 1 C .�1/ < 0C .�1/ D �1; which implies by (iv) that 0 < .�1/.�1/ D 1;a contradiction). Now define a > b to mean b < aI a � b to mean a < b or a D bI etc. Also, defineclosed interval [a; b] D fx : a � x � bgI open interval .a; b/ D fx : a < x < bgI etc. Part (i) of theorder axiom implies any two real numbers can be compared. We define max.a1; : : : ; an/ to be the maximumof a1; : : : ; an and similarly for minimum. Also, define jxj D max.x;�x/: Then x � jxj and �x � jxj;i.e. �jxj � x � jxj: Next jxj � a if and only if x � a and �x � a; i.e. �a � x � a: Finally, adding�jxj � x � jxj and�jyj � y � jyj; we get�jxj � jyj � x C y � jxj C jyj; which is the triangle inequalityjx C yj � jxj C jyj:)

(3) (Well-ordering Axiom) N D f1; 2; 3; : : :g is well-ordered (i.e. for any nonempty subset S of N, there is m 2 Ssuch that m � x for all x 2 S. This m is the least element (or the minimum) of S).

(This axiom allows us to formulate the principle of mathematical induction later.)

Definitions. For a nonempty subset S of R, S is bounded above iff there is some M 2 R such that x � M for allx 2 S. Such an M is called an upper bound of S. The supremum or least upper bound of S (denoted by sup S or lub S)is an upper bound QM of S such that QM � M for all upper bounds M of S.

(4) (Completeness Axiom) Every nonempty subset ofRwhich is bounded above has a supremum inR.

(This axiom allows us to prove results that have to do with the existence of certain numbers with specificproperties, as in the intermediate value theorem.)

Examples. (1) For S D � 1n : n 2 N D �

1; 12 ; 1

3 ; : : :; the upper bounds of S are all M � 1: So sup S D 1 2 S:(2) For S D fx 2 R: x < 0g, the upper bounds of S are all M � 0: So sup S D 0 62 S:Definitions. N D f1; 2; 3; 4; : : :g is the natural numbers (or positive integers),ZD f: : : ;�3;�2;�1; 0; 1; 2; 3; : : :gis the integers, Q D fm

n : m 2 Zand n 2 Ng is the rational numbers and Rn Q D fx 2 R: x 62 Qg is the irrationalnumbers.

Remarks (Exercises). The first three axioms are also true if R is replaced by Q: However, the completeness axiomis false for Q: For example, S D fx : x 2 Q; x > 0; x2 < 2g is bounded above by 3 in Q; but it does not have asupremum inQ:

21

As above, we define S to be bounded below if there is some m 2 R such that m � x for all x 2 S. Such an m iscalled a lower bound of S. The infimum or greatest lower bound (denoted by inf S or glb S) of S is a lower bound Qmof S such that m � Qm for all lower bounds m of S.

lower boundsare here

upper boundsare hereS

inf S sup S

Remarks (Exercises). (1) Let �B D f�x : x 2 Bg: (This is the reflection of B about 0.) If B is bounded below,then�B is bounded above and inf B D � sup.�B/: Similarly, if B is bounded above, then�B is bounded below andsup B D � inf.�B/: From these and the completeness axiom, we get the following statement.

(Completeness Axiom for Infimum) Every nonempty subset ofRwhich is bounded below has an infimum inR.

( )-B B

0inf(-B) sup(-B) supBinfA supA

AinfB

(2) For a set B, if it is bounded above and c � 0; then let cB D fcx : x 2 Bg: (This is the scaling of B by a factor ofc:) We have sup cB D c sup B: If ; 6D A � B, then inf B � inf A whenever B is bounded below and sup A � sup Bwhenever B is bounded above.

infB supB sup(c+B)inf(c+B)

c units

B c+B

(3) For c 2 R; let cC B D fc C x : x 2 Bg: (This is a translation of B by c units.) It follows that B has a supremumif and only if c C B has a supremum, in which case sup.c C B/ D c C sup B: The infimum statement is similar, i.e.inf.c C B/ D cC inf B: More generally, if A and B are bounded, then letting A C B D fx C y : x 2 A; y 2 Bg; wehave sup.A C B/ D sup A C sup B and inf.A C B/ D inf AC inf B:

If S is bounded above and below, then S is bounded. Note sup S, inf S may or may not be in S. Also, if Sis bounded, then for all x 2 S; jxj D max.x;�x/ � max.sup S;� inf S/ (because x � sup S and �x � � inf S:)Conversely, if there is c 2 R such that for all x 2 S; jxj � c; then �c � x � c so that S is bounded (above by c andbelow by �c:)Simple Consequences of the Axioms.

Theorem (Infinitesimal Principle). For x, y 2 R, x < yC " for all " > 0 if and only if x � y: (Similarly, y � " < xfor all " > 0 if and only if y � x.)

Proof. If x � y, then for all " > 0, x � y D y C 0 < y C " by (iv) of the field axiom and (iii) of the order axiom.

Conversely, if x < y C " for all " > 0; then assuming x > y; we get x � y > 0 by (iii) of the order axiom. Let"0 D x � y; then x D y C "0: Since "0 > 0; we also have x < y C "0: These contradict (i) of the order axiom. Sox � y: The other statement follows from the first statement since y � " < x is the same as y < x C ":Remarks. Taking y D 0, we see that jxj < " for all " > 0 if and only if x D 0. This is used when it is difficult toshow two expressions a; b are equal, but it may be easier to show ja � bj < " for every " > 0:Theorem (Mathematical Induction). For every n 2 N, A.n/ is a (true or false) statement such that A.1/ is true andfor every k 2 N, A.k/ is true implies A.k C 1/ is also true. Then A.n/ is true for all n 2 N.

22

Proof. Suppose A.n/ is false for some n 2 N. Then S D fn 2 N: A.n/ is falseg is a nonempty subset of N. By thewell-ordering axiom, there is a least element m in S: Then A.m/ is false. Also, if A.n/ is false, then m � n. Takingcontrapositive, this means that if n < m; then A.n/ is true.

Now A.1/ is true, so m 6D 1 and m 2 N imply m � 2. So m � 1 � 1: Let k D m � 1 2 N; then k D m � 1 < mimplies A.k/ is true. By hypothesis, A.k C 1/ D A.m/ is true, a contradiction.

Theorem (Supremum Property). If a set S has a supremum in R and " > 0; then there is x 2 S such thatsup S � " < x � sup S.

Proof. Since sup S � " < sup S, sup S � " is not an upper bound of S. Then there is x 2 S such that sup S � " < x .Since sup S is an upper bound of S, x � sup S. Therefore sup S � " < x � sup S.

Theorem (Infimum Property). If a set S has an infimum inRand " > 0; then there is x 2 S such that inf S C " >x � inf S.

Proof. Since inf SC " > inf S, inf SC " is not a lower bound of S. Then there is x 2 S such that inf SC " > x . Sinceinf S is a lower bound of S, x � inf S. Therefore inf S C " > x � inf S.

Theorem (Archimedean Principle). For any x 2 R, there is n 2 N such that n > x .

Proof. Assume there exists x 2 R such that for all n 2 N; we have n � x : Then N D fn: n 2 Ng has an upperbound x . By the completeness axiom, N has a supremum in R. By the supremum property, there is n 2 N such thatsupN� 1 < n, which yields the contradiction supN < n C 1 2 N.

Question. How isQ contained inR? How isRnQ contained inR?

Below we will show thatQ is “dense” inRin the sense that between any two distinct real numbers x; y, no matterhow close, there is a rational number. Similarly,RnQ is “dense” inR: First we need a lemma.

Lemma. For every x 2 R; there exists a least integer greater than or equal to x : (In computer science, this is calledthe ceiling of x and is denoted by dxe:) Similarly, there exists a greatest integer less than or equal to x : (This is denotedby [x]: In computer science, this is also called the floor of x and is denoted by bxc:)Proof. By the Archimedean principle, there is n 2 N such that n > jxj: Then�n < x < n: By (iii) of the order axiom,0 < x C n < 2n: The set S D fk 2 N : k � x C ng is a nonempty subset of N because 2n 2 S: By the well-orderingaxiom, there is a least positive integer m � x C n: Then m � n is the least integer greater than or equal to x : So theceiling of every real number always exist.

Next, to find the floor of x; let k be the least integer greater than or equal to �x; then �k is the greatest integerless than or equal to x :Theorem (Density of Rational Numbers). If x < y, then there is m

n 2 Q such that x < mn < y.

Proof. By the Archimedean principle, there is n 2 N such that n > 1=.y� x/: So ny�nx > 1 and hence nxC1 < ny:Let m D [nx]C 1; then m � 1 D [nx] � nx < [nx]C 1 D m: So nx < m � nx C 1 < ny; i.e. x < m

n < y:Theorem (Density of Irrational Numbers). If x < y, then there is w 2 RnQ such that x < w < y.

Proof. Let w0 2 Rn Q. By the density of rational numbers, there is mn 2 Q such that xjw0 j < m

n < yjw0 j : (If mn D 0;

then pick another rational number between 0 and yjw0 j : So we may take mn 6D 0:) Let w D m

n jw0j; then w 2 RnQ andx < w < y:Examples. (1) Let S D .�1; 3/ [ .4; 5]; then S is not bounded below and so S has no infimum. On the other hand,

S is bounded above by 5 and every upper bound of S is greater than or equal to 5 2 S: So sup S D 5:(2) Let S D f 1

n : n 2 Ng D f1; 12 ; 1

3 ; 14 ; : : :g: In the examples following the definition of supremum, we saw sup S D 1:

Here we will show inf S D 0: (Note 0 62 S.) Since 1n > 0 for all n 2 N; 0 is a lower bound of S. So by the

completeness axiom for infimum, inf S must exist. Assume S has a lower bound t > 0: By the Archimedeanprinciple, there is n 2 N such that n > 1=t : Then t > 1=n 2 S; a contradiction to t being a lower bound of S: So0 is the greatest lower bound of S:

23

(3) Let S D [2; 6/ \Q: Since 2 � x < 6 for every x 2 S; S has 2 as a lower bound and 6 as an upper bound. Wewill show inf S D 2 and sup S D 6: (Note 2 2 S and 6 62 S.) Since 2 2 S; so every lower bound t satisfy t � 2:Therefore inf S D 2: For supremum, assume there is an upper bound u < 6: Since 2 2 S; so 2 � u: By thedensity of rational numbers, there is a r 2 Q such that u < r < 6: Then r 2 [2; 6/\QD S: As u < r contradictsu being an upper bound of S; so every upper bound u � 6: Therefore, sup S D 6:

24

Chapter 6. Limits

Limit is the most important concept in analysis. We will first discuss limits of sequences, then limits of functions.

Definitions. An (infinite) sequence in a set S (e.g. S D Ror S D [0; 1]) is a list x1; x2; x3; : : : of elements of S in aspecific order. Briefly it is denoted by fxng: (Mathematically it may be viewed as a function x :N! S with x.n/ D xn

for n 2 N:) We say the sequence fxng is bounded above iff the set fx1; x2; x3; : : :g is bounded above. (Bounded belowand bounded sequences are defined similarly.) We will also write supfxng for the supremum of the set fx1; x2; x3; : : :gand inffxng for the infimum of the set fx1; x2; x3; : : :g:CAUTION: Since we seldom talk about a set with one element from now on, so notations like fxng will denotesequences unless explicitly stated otherwise.

For x; y 2 R; the distance between x and y is commonly denoted by d.x; y/; which equals jx� yj: Below we willneed a quantitative measure of what it means to be “close” for a discussion of the concept of limit. For " > 0; the openinterval .c� "; cC "/ is called the "-neighborhood of c: Note x 2 .c � "; cC "/ if and only if d.x; c/ D jx � cj < ";i.e. every number in .c � "; cC "/ has distance less than " from c:

Limit of a sequence fxng is often explained by saying it is the number the xn’s are closer and closer to as n getslarger and larger. There are two bad points about this explanations.

(1) Being close or large is a feeling! It is not a fact. It cannot be proved by a logical argument.

(2) The effect of being close can accumulate to yield large separation! If two numbers having a distance less than orequal to 1 are considered close, then 0 is close to 1 and 1 is close to 2 and 2 is close to 3, : : : ; 99 is close to 100,but 0 is quite far from 100.

So what is the meaning of close? How can limit be defined so it can be checked? Intuitively, a sequence fxng getsclose to a number x if and only if the distance d.xn; x/ goes to 0. This happens if and only if for every positive "; thedistance d.xn; x/ eventually becomes less than ": The following example will try to make this more precise.

Example. As n gets large, intuitively we may think xn D 2n2 � 1

n2 C 1gets close to 2. For " D 0:1; how soon (that is, for

what n) will the distance d.xn; 2/ be less than "? (What if " D 0:01? What if " D 0:001? What if " is an arbitrarypositive number?)

Solution. Consider d.xn; 2/ D ����2n2 � 1n2 C 1

� 2

���� D 3n2 C 1

< ": Solving for n; we get n2 > .3="/� 1: If " D 0:1; then

n > p29: So as soon as n � 6; the distance between xn and 2 will be less than " D 0:1:

(If " D 0:01; then n > p299: So n � 18 will do. If " D 0:001; then n > p

2999: So n � 55 will do.If 0 < " � 3; then n � [

p.3="/ � 1] C 1 will do. If " > 3; then since 3n2C1 < 3 < " for every n 2 N; so

n � 1 will do. So for every " > 0; there is a K 2 N so that as soon as n � K ; the distance d.xn; 2/ will be lessthan ":) Note the value of K depends on the value of "I the smaller " is, the larger K will be. (Some people write K"to indicate K depends on ":)Definition. A sequence fxng converges to a number x (or has limit x) iff for every " > 0, there is K 2 N such that forevery n � K ; it implies d.xn; x/ D jxn � xj < " (which means xK ; xKC1; xKC2; : : : 2 .x � "; x C "/:/Remarks. (i) From the definition, we see that fxng converges to x; fxn � xg converges to 0 and fjxn � xjg converges

to 0 are equivalent because in the definition, jxn � xj is the same as j.xn � x/ � 0j D jjxn � xj � 0j:(ii) To show fxng converges to x means for every " > 0; we have to find a K as in the definition or show such a K

exists. On the other hand, if we are given that fxng converges to x , then for every " > 0; (which we can evenchoose for our convenience,) there is a K as in the definition for us to use.

Let us now do a few more examples to illustrate how to show a sequence converges by checking the definition.Later, we will prove some theorems that will help in establishing convergence of sequences.

25

Examples. (1) Let vn D c: For every " > 0, let K D 1, then n � K implies jvn � cj D 0 < ". So fvng converges to c:(2) Let wn D c� 1

n: For every " > 0, there exists an integer K > 1" (by the Archimedean principle). Then n � K

implies jwn � cj D 1

n� 1

K< ". So fwng converges to c:

(3) Let xn D n.cos n/ � n: Show that fxng converges to�1 by checking the definition.

Solution. For every " > 0; there exists an integer K > 1C 1" by the Archimedian principle. Then n � K implies�� n.cos n/ � n� .�1/�� D �� cos n.cos n/� n

�� � 1

n � 1� 1

K � 1< ": So fxng converges to �1:

(4) Let yn D .�1/n : Show that fyng does not converge.

Solution. Assume fyng converges, say to y: Let " D 0:1: Then there exists K 2 N such that n � K impliesj.�1/n � yj < " D 0:1: Taking an odd integer n � k; we get j � 1 � yj < 0:1; which implies y 2 .�1:1;�0:9/:Taking a even integer n � K ; we get j1� yj < 0:1; which implies y 2 .0:9; 1:1/: Since no y is in both .�1:1;�0:9/and .0:9; 1:1/; we have a contradiction.

(5) Let zn D n1=n: Show that fzng converges to 1 by checking the definition.

Solution. (Let un D jzn � 1j D zn � 1: By the binomial theorem,

n D znn D .1C un/n D 1C nun C n.n � 1/

2u2

n C � � � C unn � n.n � 1/

2u2

n

so that un � r 2

n � 1:) For every " > 0; there exists integer K > 1C 2"2

(by the Archimedean principle). Then n � K

implies jzn � 1j D un � r 2

n � 1� r 2

K � 1< ": So fzng converges to 1.

Theorem (Uniqueness of Limit). If fxng converges to x and y, then x D y (and so we may write limn!1 xn D x).

Proof. For every " > 0, we will show jx�yj < ". (By the infinitesimal principle,we will get x D y.) Let "0 D "=2 > 0:By the definitionof convergence, there are K1, K2 2 N such that n � K1 ) jxn�xj < "0 and n � K2 ) jxn� yj < "0.Let K D max.K1; K2/. By the triangle inequality, jx� yj D j.x�xK /C.xK � y/j � jx�xK jCjxK � yj < "0C"0 D ".

Boundedness Theorem. If fxng converges, then fxng is bounded.

Proof. Let limn!1 xn D x . For " D 1, there is K 2 N such that n � K ) jxn�xj < 1 ) jxnj D j.xn�x/Cxj < 1Cjxj.

Let M D max.jx1j, : : :, jxK�1j, 1C jxj/; then for every n 2 N; jxnj � M (i.e. xn 2 [�M; M]).

Remarks. The converse is false. The sequence f.�1/n g is bounded, but not convergent by example (4). In general,bounded sequences may or may not converge.

Theorem (Computation Formulas for Limits). If fxng converges to x and fyng converges to y, then

(i) fxn � yng converges to x � y, respectively, i.e. limn!1.xn � yn/ D lim

n!1 xn � limn!1 yn,

(ii) fxn yng converges to x y, i.e. limn!1.xn yn/ D �

limn!1 xn

��lim

n!1 yn

�,

(iii) fxn=yng converges to x=y, provided yn 6D 0 for all n and y 6D 0.

26

Proof. (i) For every " > 0, there are K1; K2 2 N such that n � K1 ) jxn � xj < "=2 and n � K2 ) jyn � yj < "=2.Let K D max.K1; K2/: Then n � K implies n � K1 and n � K2: So for these n’s,j.xn � yn/ � .x � y/j D j.xn � x/ � .yn � y/j � jxn � xj C jyn � yj < "=2C "=2 D ":(ii) We prove a lemma first.

Lemma. If fang is bounded and limn!1 bn D 0, then lim

n!1 anbn D 0.

Proof. Since fang is bounded, there is M such that janj < M for all n. For every " > 0, since "=M > 0 andfbng converges to 0, there is K 2 N such that n � K ) jbn � 0j < "=M ) janbn � 0j � Mjbnj < ".

To prove (ii), we write xn yn � x y D xn yn � xn y C xn y � x y D xn.yn � y/ C y.xn � x/. Since fxng converges,fxng is bounded by the boundedness theorem. So by (i) and the lemma,

limn!1 xn yn D lim

n!1.xn yn � x y/C limn!1 x y D lim

n!1 xn.yn � y/ C limn!1 y.xn � x/ C x y D 0C 0C x y D x y:

(iii) Note 12 jyj > 0: Since fyng converges to y; there is K0 2 N such that n � K0 implies jyn � yj < 1

2 jyj: Bythe triangle inequality, jyj � jynj � jyn � yj < 1

2 jyj ) 12 jyj < jynj for n � K0: Then for every n 2 N;jynj � m D min.jy1j; : : : ; jyK0�1j; 1

2 jyj/ > 0:Next we will show lim

n!1 1ynD 1

y(then by (ii), lim

n!1 xn

ynD lim

n!1.xn1yn/ D x

1yD x

y). For every " > 0; let"0 D mjyj" > 0: Since lim

n!1 yn D y 6D 0, there is K 2 N such that n � K ) jyn � yj < "0: Then

n � K ) ���� 1yn� 1

y

���� D jy � ynjjynjjyj < "0

mjyj D ":Remarks. (1) As in the proofs of the uniqueness of limit and part (i) of the computation formulas, when wehave n � K1 ) jan � aj < "1 and n � K2 ) jbn � bj < "2; we may as well take K D max.K1; K2/ to sayn � K ) jan � aj < "1 and jbn � bj < "2 from now on.

(2) By mathematical induction, we can show that the computation formulas also hold for finitely many sequences.However, the number of sequences must stay constant as the following example shows

1 D limn!1�1

nC � � � C 1

n| {z }n terms

� 6D limn!1 1

nC � � � C lim

n!1 1nD 0C � � � C 0 D 0:

Sandwich Theorem (or Squeeze Limit Theorem). If xn � wn � yn for all n 2 N and limn!1 xn D lim

n!1 yn D z; then

limn!1wn D z.

Proof. For any " > 0, there is K such that n � K ) jxn � zj < " and jyn � zj < ", i.e. xn , yn 2 .z � "; zC "/. Sincexn � wn � yn; so wn 2 .z � "; z C "/, i.e. jwn � zj < ".

Example. Let wn D [10np

2]

10n2 Q for every n 2 N: (Note w1 D 1:4; w2 D 1:41; w3 D 1:414; w4 D 1:4142; : : : :)

Then10n

p2� 1

10n< wn � 10n

p2

10nD p

2: Since limn!1 10n

p2� 1

10nD p

2; by the sandwich theorem, limn!1wn D p

2:Theorem (Limit Inequality). If an � 0 for all n 2 N and lim

n!1 an D a, then a � 0.

Proof. Assume a < 0. Then for " D jaj, there is K 2 N such that n � K ) jan � aj < " D jaj, which impliesa � " < an < a C " D a C .�a/ D 0, a contradiction.

Remarks. By the limit inequality above, if xn � yn , limn!1 xn D x , lim

n!1 yn D y, then taking an D yn � xn � 0; we get

the limit y � x � 0; i.e. x � y: Also, if a � xn � b and limn!1 xn D x; then lim

n!1 a D a � limn!1 xn D x � lim

n!1 b D b;i.e. xn 2 [a; b] and lim

n!1 xn D x imply x 2 [a; b]: (This is false for open intervals as 1n 2 .0; 2/; lim

n!1 1

nD 0 62 .0; 2/:)

27

Supremum Limit Theorem. Let S be a nonempty set with an upper bound c: There is a sequence fwng in S convergingto c if and only if c D sup S.

Proof. If c D sup S; then for n 2 N, by the supremum property, there is wn 2 S such that c� 1n D sup S � 1

n < wn �sup S D c. Since lim

n!1.c � 1

n/ D lim

n!1 c D c, the sandwich theorem implies limn!1wn D c D sup S.

Conversely, if a sequence fwng in S converges to c; then wn � sup S implies c D limn!1wn � sup S: Since c is an

upper bound of S; so sup S � c: Therefore c D sup S:Infimum Limit Theorem. Let S be a nonempty set with a lower bound c: There is a sequence fwng in S convergingto c if and only if c D inf S.

Examples. (1) Let S D f 1n : n 2 Ng D f1; 1

2 ; 13 ; : : :g: Since 0 � 1

n for all n 2 N; 0 is a lower bound of the set S: Nowthe sequence f 1

n g in S converges to 0: By the infimum limit theorem, inf S D 0:(2) Let S D fx� C 1

y : x 2 Q\ .0; 1]; y 2 [1; 2]g: Since 12 � x� C 1

y for all x 2 Q\ .0; 1] and y 2 [1; 2]; 12 is a

lower bound of S: Now the sequence f 1n� C 1

2 g in S converges to 12 : By the infimum limit theorem, inf S D 1

2 :(3) Let A and B be bounded inR: Prove that if A�2B D fa�2b : a 2 A; b 2 Bg; then sup.A�2B/ D sup A�2 inf B:

Solution. Since A and B are bounded, sup A and inf B exist by the completeness axiom. For x 2 A � 2B; wehave x D a�2b for some a 2 A and b 2 B: So x D a�2b � sup A�2 inf B: Hence sup A�2 inf B is an upperbound for A�2B: By the supremum limit theorem, there is a sequence an 2 A such that fang converges to sup A:By the infimum limit theorem, there is a sequence bn 2 B such that fbng converges to inf B: Then fan � 2bng is asequence in A� 2B and fan � 2bng converges to sup A� 2 inf B by the computation formulas for limits. By thesupremum limit theorem, therefore sup.A � 2B/ D sup A � 2 inf B:

Definition. A subsequence of fxng is a sequence fxnj g, where n j 2 N and n1 < n2 < n3 < : : :.Examples. For the sequence x1; x2; x3; x4; x5; x6; : : : ; if we set n j D j 2; we get the subsequence x1; x4; x9; x16; : : : :If n j D 2 j C 1; then we get the subsequence x3; x5; x7; x9; : : : : If n j is the j -th prime number, then we get thesubsequence x2; x3; x5; x7; : : : :Remarks. (1) Taking n j D j; we see that every sequence is a subsequence of itself. A subsequence can also bethought of as obtained from the original sequence by throwing away possibly some terms. Also, a subsequence of asubsequence of fxng is a subsequence of fxng:(2) By mathematical induction, we have n j � j for all j 2 N because n1 � 1 and n jC1 > n j � j implies n jC1 � j C1:Subsequence Theorem. If lim

n!1 xn D x, then limj!1 xnj D x for every subsequence fxnj g of fxng. (The converse is

trivially true because every sequence is a subsequence of itself.)

Proof. For every " > 0, there is K 2 N such that n � K ) jxn � xj < ". Then j � K ) n j � K ) jxnj � xj < ".

Question. How can we tell if a sequence converges without knowing the limit (especially if the sequence is given bya recurrence relation)?

For certain types of sequences, the question has an easy answer.

Definitions. fxng is

8><>: increasingdecreasing

strictly increasingstrictly decreasing

9>=>; iff

8><>: x1 � x2 � x3 � : : :x1 � x2 � x3 � : : :x1 < x2 < x3 < : : :x1 > x2 > x3 > : : :9>=>; ; respectively. fxng is

�monotone

strictly monotone

�ifffxng is

�increasing or decreasing

strictly increasing or decreasing

� ; respectively.

Monotone Sequence Theorem. If fxng is increasing and bounded above, then limn!1 xn D supfxng. (Similarly, if fxng

is decreasing and bounded below, then limn!1 xn D inffxng.)

28

Proof. Let M D supfxng, which exists by the completeness axiom. By the supremum property, for any " > 0, there isxK such that M � " < xK � M . Then j � K ) M � " < xK � x j � M ) jx j � Mj D M � x j < ":Remark. Note the completeness axiom was used to show the limit of xn exists (without giving the value).

Examples. (1) Let 0 < c < 1 and xn D c1=n: Then xn < 1 and cnC1 < cn ) xn D c1=n < c1=.nC1/ D xnC1: So bythe monotone sequence theorem, fxng has a limit x : Now x2

2n D .c1=2n/2 D c1=n D xn: Taking limits and using thesubsequence theorem, we get x2 D x : So x D 0 or 1. Since 0 < c D x1 � x; the limit x is 1. Similarly, if c � 1; thenc1=n will decrease to the limit 1.

(2) Does

r2Cq2Cp

2C � � � represent a real number?

Here we have a nested radical defined by x1 D p2 and xnC1 D p

2C xn: The question is whether fxngconverges to a real number x . (Computing a few terms, we suspect that fxng is increasing. To find an upper bound,observe that if lim

n!1 xn D x; then x D p2C x implies x D 2:) Now by mathematical induction, we can show that

xn < xnC1 < 2: (If xn < xnC1 < 2; then 2C xn < 2C xnC1 < 4; so taking square roots, we get xnC1 < xnC2 < 2:)By the monotone sequence theorem, fxng has a limit x : We have x2 D lim

n!1 x2nC1 D lim

n!1 2C xn D 2C x : Then

x D �1 or 2. Sincep

2 D x1 � x; so x D 2:Another common type of sequences is obtained by mixing a decreasing sequence and an increasing sequence into

one of the form a1; b1; a2; b2; a3; b3; : : : : In the next example, we will have such a situation and we need two theoremsto handle these kind of sequences.

Nested Interval Theorem. If In D [an; bn] is such that I1 � I2 � I3 � : : :, then1\

nD1In D [a; b]; where

a D limn!1 an � lim

n!1 bn D b: If limn!1.bn � an/ D 0, then

1\nD1

In contains exactly one number.

[ ][ [ ] ]a1 a2 a3 b3 b2 b1...

Proof. I1 � I2 � I3 � : : : implies fang is increasing and bounded above by b1 and fbng is decreasing and boundedbelow by a1. By the monotone sequence theorem, fang converges to a D supfang and fbng converges to b D inffbng.Since an � bn for every n 2 N, taking limits, we have an � a � b � bn . Consequently, x 2 [an; bn] (i.e. an � x � bn)

for all n if and only if limn!1 an D a � x � b D lim

n!1 bn: So1\

nD1In D [a; b]. If 0 D lim

n!1.bn � an/ D b� a; then a D b

and1\

nD1In D fag:

Remarks. Note in the proof, the monotone sequence theorem was used. So the nested interval theorem also implicitlydepended on the completeness axiom.

Intertwining Sequence Theorem. If fx2mg and fx2m�1g converge to x; then fxng also converges to x :Proof. For every " > 0; since fx2mg converges to x; there is K0 2 N such that m � K0 ) jx2m� xj < ": Since fx2m�1galso converges to x; there is K1 2 N such that m � K1 ) jx2m�1 � xj < ": Now if n � K D max.2K0; 2K1 � 1/;then either n D 2m � 2K0 ) jxn � xj D jx2m � xj < " or n D 2m � 1 � 2K1 � 1 ) jxn � xj D jx2m�1 � xj < ":Example. Does

1

1C 1

1C � � � represent a number?

Here we have a continued fraction defined by x1 D 1 and xnC1 D 1=.1 C xn/: We have x1 D 1; x2 D 1=2; x3 D2=3; x4 D 3=5; : : : : Plotting these on the real line suggests 1=2 � x2n < x2nC2 < x2nC1 < x2n�1 � 1 for all n 2 N:This can be easily established by mathematical induction. (If 1=2 � x2n < x2nC2 < x2nC1 < x2n�1 � 1; then1 C x2n < 1 C x2nC2 < 1 C x2nC1 < 1 C x2n�1: Taking reciprocal and applying the recurrence relation, we havex2nC1 > x2nC3 > x2nC2 > x2n : Repeating these steps once more, we get x2nC2 < x2nC4 < x2nC3 < x2nC1:)

Let In D [x2n; x2n�1]; then I1 � I2 � I3 � � � � : Nowjxm � xmC1j D ���� 1

1C xm�1� 1

1C xm

���� D jxm�1 � xm j.1C xm�1/.1 C xm/ < jxm�1 � xmj.1C 12/.1 C 1

2/ D 4

9jxm�1 � xm j:

29

Using this, we get jx2n�1 � x2nj < 4

9jx2n�2 � x2n�1j < 4

9

4

9jx2n�3 � x2n�2j < � � � < 4

9� � � 4

9| {z }2n�2

jx1 � x2j D �4

9

�2n�2 1

2:

By the sandwich theorem, limn!1.x2n�1 � x2n/ D 0: So by the nested interval theorem,

1\nD1

In D fxg for some

x and limn!1 x2n D x D lim

n!1 x2n�1: By the intertwining sequence theorem, limn!1 xn D x : So x D lim

n!1 xnC1 Dlim

n!1 1=.1C xn/ D 1=.1C x/: Then x D .�1�p5/=2: Since x 2 I1; x D .�1Cp

5/=2:(Instead of estimating the lengths of the In’s and squeezing them to 0 to see their intersection is a single point,

we can also do the following. Let In D [x2n; x2n�1] be as above so that I1 � I2 � I3 � � � � : By the nested interval

theorem,1\

nD1In D [a; b];where a D lim

n!1 x2n and b D limn!1 x2n�1: Taking limits on both sides of x2nC1 D 1

1C x2nand

x2n D 11C x2n�1

; we get b D 11C a

and a D 11C b

: Then b.1C a/ D 1 D a.1C b/; which yields b C ab D a C ab;so a D b: Hence

1\nD1

In is a single point.)

Back to answering the question above in general, French mathematician Augustine Cauchy (1789–1857) intro-duced the following condition.

Definition. fxng is a Cauchy sequence iff for every " > 0, there is K 2 N such that m, n � K implies jxm � xnj < ".

Remark. Roughly, the condition says that the terms of these sequences are getting closer and closer to each other.

Example. Let xn D 1

n2: (Note that if m; n � K ; say m � n; then we have jxm � xnj D 1

n2� 1

m2< 1

n2� 1

K 2:)

For every " > 0; we can take an integer K > 1p" (by the Archimedean principle). Then m; n � K impliesjxm � xnj � 1K 2

< ": So fxng is a Cauchy sequence.

Theorem. If fxng converges, then fxng is a Cauchy sequence.

Proof. For every " > 0; since limn!1 xn D x; there is K 2 N such that j � K ) jx j � xj < "=2. For m, n � K , we

have jxm � xnj � jxm � xj C jx � xnj < "=2C "=2 D ". So fxng is a Cauchy sequence.

The converse of the previous theorem is true, but it takes some work to prove that. The difficulty lies primarilyon how to come up with a limit of the sequence. The strategy of showing every Cauchy sequence inRmust convergeis first to find a subsequence that converges, then show that the original sequence also converge to the same limit.

Theorem. If fxng is a Cauchy sequence, then fxng is bounded.

Proof. Let " D 1: Since fxng is a Cauchy sequence, there is K 2 N such that m; n � K ) jxm � xnj < " D 1:In particular, for n � K ; jxK � xnj < 1 ) jxnj D j.xn � xK / C xK j � jxn � xK j C jxK j < 1 C jxK j: LetM D max.jx1j; : : : ; jxK�1j; 1C jxK j/; then for all n 2 N; jxnj � M (i.e. xn 2 [�M; M]/:Bolzano-Weierstrass Theorem. If fxng is bounded, then fxng has a subsequence fxnj g that converges.

Proof. (Bisection Method) Let a1 D inffxng, b1 D supfxng and I1 D [a1; b1]. Let m1 be the midpoint of I1. If thereare infinitely many terms of fxng in [a;m1], then let a2 D a1, b2 D m1 and I2 D [a2; b2]. Otherwise, there will beinfinitely many terms of fxng in [m1; b1], then let a2 D m1, b2 D b1 and I2 D [a2; b2]. For k D 2, 3, 4, : : :, repeatthis bisection on Ik to get IkC1. We have Ij D [aj; bj] and I1 � I2 � I3 � : : :. By the nested interval theorem, since

limj!1.bj � aj/ D lim

j!1 b1 � a1

2 j�1D 0,

1\nD1

In contains exactly one number x .

Take n1 D 1; then xn1 D x1 2 I1. Suppose n j is chosen with xnj 2 Ij . Since there are infinitely many terms xn inIjC1, choose n jC1 > n j and xnjC1 2 IjC1. Then lim

j!1 jxnj � xj � limj!1.bj � aj/ D 0. Therefore, lim

j!1 xnj D x .

30

Remarks. In the proof, the nested interval theorem was used, so the Bolzano-Weierstrass theorem depended on thecompleteness axiom.

Alternate Proof. We will show every sequence fxng has a monotone subsequence. (If fxng is bounded, then themonotone sequence theorem will imply the subsequence converges.)

Call xm a peak of fxng if xm � xk for all k > m: If fxng has infinitely many peaks, then we order the peaks bystrictly increasing subscripts m1 < m2 < m3 < � � � : By the definition of a peak, xm1 � xm2 � xm3 � � � � : So fxm j gis a decreasing subsequence of fxng: On the other hand, if fxng has only finitely many peaks xm1; � � � ; xmk; then letn1 D maxfm1; � � � ;mkg C 1: Since xn1 is not a peak, there is n2 > n1 such that xn2 > xn1 : Inductively, if xnj is not apeak, there is n jC1 > n j with xnjC1 > xnj : So fxnj g is a strictly increasing subsequence of fxng:Remarks. This alternate proof used the monotone sequence theorem, so it also depended on the completeness axiom.

Cauchy’s Theorem. fxng converges if and only if fxng is a Cauchy sequence.

Proof. The ‘only if’ part was proved. For the ‘if’ part, since fxng is a Cauchy sequence, fxng is bounded. By theBolzano-Weierstrass theorem, fxng has a subsequence fxnj g that converges inR, say lim

j!1 xnj D x .

We will show limn!1 xn D x . For every " > 0, since fxng is a Cauchy sequence, there is K1 2 N such that m,

n � K1 ) jxm � xnj < "=2. Since limj!1 xnj D x , there is K2 2 N such that j � K2 ) jxnj � xj < "=2. If

n � J D max.K1; K2/, then n J � J � K1, J � K2 and jxn � xj � jxn � xnJ j C jxnJ � xj < "=2C "=2 D ".

Example. Does the sequence fxng converge, where x1 D sin 1 and xk D xk�1 C sin k

k2for k D 2; 3; 4; : : :?

We will check the Cauchy condition. For m > n; xm � xn D mXkDnC1

.xk � xk�1/ D mXkDnC1

sin k

k2andjxm�xnj � 1.n C 1/2

C� � �C 1

m2< 1

n.n C 1/C� � �C 1.m � 1/m D �1

n� 1

n C 1

�C� � �C� 1

m � 1� 1

m

�D 1

n� 1

m< 1

n:

So for " > 0; by the Archimedean principle, there is K > 1=": Then m; n � K ) jxm � xnj < 1K < ": Therefore, by

the Cauchy theorem, fxng converges.

Limits of Functions

Let S be an interval (or more generally a set). Suppose f : S ! Ris a function. We would like to define limx!x0

f .x/.For this to be a meaningful expression, in the interval case, x0 must be a point on the interval or an endpoint. In thegeneral case, x0 must be “approachable” by the points of S.

CONVENTION. In discussing the limit of a function f : S ! Rat x0; x0 is always assumed to be the limit of somesequence in S n fx0g so that x0 can be approached by points of S: (We say x0 is an accumulation (or limit or cluster)point of S iff x0 is the limit of a sequence fxng in S n fx0g:)

(x

L

S0

y = f x)(

()x0S S

y = f (x)

LLet f : S ! R be a function. To say lim

x!x0f .x/ D L

roughly means the distance between f .x/ and L is as smallas we please when x 2 S is sufficiently close to x0.(Again,small and close need to be clarified.)

Example. Let f .x/ D .x3 � 3x2/=.x � 3/: If x gets close to 3 and x 6D 3; then f .x/ D x2 should be close to 9. Inother words, the distance d

�f .x/; 9

� D j f .x/ � 9j goes to 0 when the distance d.x; 3/ D jx � 3j gets small. So for

31

every positive "; the distance j f .x/ � 9j will soon or later be less than " when the distance jx � 3j becomes smallenough. For " D 0:1; how small d.x; 3/ be (that is, for what x) will make d

�f .x/; 9

� D j f .x/�9j < "? Equivalently,we are seeking a positive � so that 0 < jx � 3j < � ) j f .x/ � 9j < ":

Now j f .x/ � 9j < " , 8:9 < f .x/ < 9:1: So 2:9833 � p8:9 < x < p

9:1 � 3:0166 and �0:0167 < x � 3 <0:0166: If we take � D 0:016; then 0 < jx � 3j < � ) j f .x/ � 9j < ":

(For small " > 0; j f .x/ � 9j < ", 9� " < f .x/ < 9C " ) p9� "� 3 < x � 3 < p

9C " � 3: So we maytake � D min.p9C " � 3; 3�p

9� "/; then 0 < jx � 3j < � ) j f .x/ � 9j < ":)Following the example, we are ready to state the precise definition of the limit of a function.

Definition. Let f : S ! Rbe a function. We say f .x/ converges to L (or has limit L) as x tends to x0 in S iff forevery " > 0, there is � > 0 such that for every x 2 S; 0 < jx � x0j < � implies j f .x/ � L j < ". This is denoted bylim

x!x0x2S

f .x/ D L (or limx!x0

f .x/ D L in short.)

In the definition, � depends on " and x0: For different " (or different x0), � will be different. If a limit value exists,then it is unique. The proof is similar to the sequential case and is left as an exercise for the readers.

Examples. (1) For g : [0;1/ ! Rdefined by g.x/ D px; show that lim

x!0g.x/ D 0 and lim

x!4g.x/ D 2 by checking

the "-� definition.

Solution. For every " > 0; let � D "2: Then for every x 2 [0;1/; 0 < jx�0j < � implies jg.x/�0j D px < p� D ":

This takes care of the checking for the first limit.

For the second limit, note that jpx � 2j D jx � 4jpx C 2

� jx � 4j2

: For every " > 0; let � D 2": Then for every

x 2 [0;1/; 0 < jx � 4j < � implies jg.x/ � 2j � jx � 4j2

< �2D ":

(2) Let f : Rn f0g ! Rbe defined by f .x/ D 15x

: Show that limx!2

f .x/ D 110

by checking the "-� definition.

Solution. (Note that

���� 1

5x� 1

10

���� D jx � 2j10x

� jx � 2j10

for x 2 .1; 3/:) For every " > 0; take � D min.1; 10"/: Then� � 1 and � � 10": For every x 2 Rn f0g; 0 < jx � 2j < � implies x 2 .1; 3/: So

���� f .x/ � 1

10

���� � jx � 2j10

< �10

� "and we are done.

Notation: We will write xn ! x0 in S n fx0g to mean sequence fxng in S n fx0g converges to x0:Sequential Limit Theorem. lim

x!x0x2S

f .x/ D L if and only if for every xn ! x0 in S n fx0g, limn!1 f .xn/ D L :

In quantifier symbols (8 D for any, for every, for all, 9 D there is, there exists),

limx!x0x2S

f .x/ D L () 8" > 0; 9� > 0 such that 8x 2 S; 0 < jx � x0j < � ) j f .x/ � L j < ";lim

x!x0x2S

f .x/ 6D L () 9" > 0 such that 8� > 0; 9x 2 S; 0 < jx � x0j < � and j f .x/ � L j � ":Proof. If lim

x!x0x2S

f .x/ D L ; then for every " > 0, there is � > 0 such that for x 2 S, 0 < jx�x0j < � ) j f .x/�L j < ".

If limn!1 xn D x0 and xn 6D x0, then there is K 2 N such that n � K ) 0 < jxn � x0j < �.) j f .xn/ � L j < "/. So

limn!1 f .xn/ D L .

32

Conversely, if limx!x0x2S

f .x/ 6D L ; then there is " > 0 such that for every � > 0; there is x 2 S with 0 < jx� x0j < �and j f .x/ � L j � ". Now, setting � D 1

n ; there is xn 2 S with 0 < jxn � x0j < � D 1n and j f .xn/ � L j � ". By the

sandwich theorem, xn ! x0 in S n fx0g: So limn!1 f .xn/ D L : Then 0 D lim

n!1 j f .xn/ � L j � ", a contradiction.

Remarks. If limn!1 f .xn/ exists for every xn ! x0 in S nfx0g; then all the limit values are the same. To see this, suppose

xn ! x0 and wn ! x0 in S n fx0g: Then the intertwining sequence fzng D fx1; w1; x2; w2; x3; w3; : : :g converges to x0

in S n fx0g: Since f f .xn/g and f f .wn/g are subsequences of the convergent sequence f f .zn/g, they have the same limit.

Consequences of Sequential Limit Theorem.

(1) (Computation Formulas) If f; g : S ! Rare functions, limx!x0x2S

f .x/ D L1 and limx!x0x2S

g.x/ D L2; then

limx!x0x2S

�f .x/8><>:C��= 9>=>; g.x/� D L1

8><>:C��= 9>=>; L2 D limx!x0x2S

f .x/8><>:C��= 9>=>; limx!x0x2S

g.x/respectively (in the case of division, provided g.x/ 6D 0 for every x 2 S and L2 6D 0).

Proof. Since f .x/ and g.x/ have limits L1 and L2; respectively, as x tends to x0 in S; by the sequential limittheorem, f .xn/ and g.xn/ will have limits L1 and L2; respectively, for every xn ! x0 in S n fx0g: By thecomputation formulas for sequences, the limit of f .xn/C g.xn/ is L1 C L2 for every xn ! x0 in S n fx0g: By thesequential limit theorem, f .x/ C g.x/ has limit L1 C L2 as x tends to x0 in S: ReplacingC by�; �; =;we get theproofs for the other parts.

Alternate Proof. If limx!x0x2S

f .x/ D L1 and limx!x0x2S

g.x/ D L2; then for every " > 0; there are �1 > 0 such that for

every x 2 S; 0 < jx� x0j < �1 implies j f .x/ � L1j < "2

and �2 > 0 such that for every x 2 S; 0 < jx� x0j < �2

implies jg.x/ � L2j < "2: Now we take � D min.�1; �2/ > 0 so that � � �1 and � � �2: Then, for every x 2 S;

0 < jx � x0j < � implies both 0 < jx � x0j < �1 and 0 < jx � x0j < �2 so that��� f .x/ C g.x/� � .L1 C L2/�� D j. f .x/ � L1/ C .g.x/ � L2/j � j f .x/ � L1j C jg.x/� L2j < "2C "

2D ":

The other parts of the computation formulas can be proved by adapting the arguments for the sequential case.

(2) (Sandwich Theorem or Squeeze Limit Theorem) If f .x/ � g.x/ � h.x/ for every x 2 S and limx!x0x2S

f .x/ D L Dlim

x!x0x2S

h.x/; then limx!x0x2S

g.x/ D L :(3) (Limit Inequality) If f .x/ � 0 for all x 2 S and lim

x!x0x2S

f .x/ D L ; then L � 0:The proofs of (2) and (3) can be done by switching to sequences like the first proof of (1) or by adapting the

arguments of the sequential cases and checking the "-� definition like the alternate proof of (1).

Next we will discuss one-sided limits.

Definitions. For f : .a; b/! Rand x0 2 .a; b/; the left hand limit of f at x0 is f .x0�/ D limx!x�0 f .x/ D lim

x!x0x2.a;x0/ f .x/.

The right hand limit of f at x0 is f .x0C/ D limx!xC0 f .x/ D lim

x!x0x2.x0;b/ f .x/:

Theorem. For x0 2 .a; b/; limx!x0

x2.a;b/ f .x/ D L if and only if f .x0�/ D L D f .x0C/.33

Proof. If limx!x0

f .x/ D L ; then for every " > 0; there is � > 0 such that for x 2 .a; b/; 0 < jx � x0j < � )j f .x/ � L j < ": In particular, for x 2 .a; x0/; 0 < jx � x0j < � ) j f .x/ � L j < ": So f .x0�/ D L : Similarly, forx 2 .x0; b/; 0 < jx � x0j < � ) j f .x/ � L j < ": So f .x0C/ D L :

Conversely, if f .x0�/ D L D f .x0C/; then for every " > 0; there is �1 > 0 such that for x 2 .a; x0/;0 < jx�x0j < �1 ) j f .x/�L j < " and there is �2 > 0 such that for x 2 .x0; b/; 0 < jx�x0j < �2 ) j f .x/�L j < ":As � D min.�1; �2/ � �1 and �2; we have for x 2 .a; b/; 0 < jx � x0j < � ) j f .x/ � L j < ": So lim

x!x0f .x/ D L :

Definitions. A function f : S ! Ris

8><>: increasingdecreasing

strictly increasingstrictly decreasing

9>=>; on S iff for every x , y 2 S, x < y ) 8><>: f .x/ � f .y/f .x/ � f .y/f .x/ < f .y/f .x/ > f .y/9>=>; :

Also, f is�

monotonestrictly monotone

�on S iff f is

�increasing or decreasing

strictly increasing or strictly decreasing

�on S; respectively.

For a nonempty subset S0 of S; we say f is

( bounded abovebounded below

bounded

)on S0 iff f f .x/ : x 2 S0g is

( bounded abovebounded below

bounded

) ;respectively. If The set S0 is not mentioned, then it is the domain S:

For monotone functions, the following theorem is analogous to the monotone sequence theorem. It will be used

in the next chapter to prove the continuous inverse theorem, which will be used to prove thedx

dyD 1

�dy

dxrule later.

Also, it will be used again in the chapter on integration.

Monotone Function Theorem. If f is increasing on .a; b/, then for every x0 2 .a; b/, f .x0�/ D supf f .x/ : a <x < x0g and f .x0C/ D inff f .x/ : x0 < x < bg and f .x0�/ � f .x0/ � f .x0C/. If f is bounded below, thenf .aC/ D inff f .x/ : a < x < bg: If f is bounded above, then f .b�/ D supf f .x/ : a < x < bg: Also f has countablymany discontinuous points on .a; b/; i.e. J D fx0 : x0 2 .a; b/; f .x0�/ 6D f .x0C/g is countable. (The theorem issimilarly true for decreasing functions and all other kinds of intervals.)

Proof. If a < x < x0 < b, then f .x/ � f .x0/. So M D supf f .x/ : a < x < x0g � f .x0/: By the supremumproperty, for every " > 0; there is c 2 .a; x0/ such that M � " < f .c/ � M: If we let � D x0 � c; then8x 2 .a; x0/; 0 < jx � x0j < � ) c D x0 � � < x < x0 ) f .c/ � f .x/ � M ) j f .x/ � Mj � M � f .c/ < ":So lim

x!x�0 f .x/ D M � f .x0/. Similarly, f .x0/ � limx!xC0 f .x/ D inff f .x/ : x0 < x < bg. In the case f is bounded

below or above, the proof of the existence of f .aC/ or f .b�/ is similar.

q (x1)

q(x0)

x x0 1

Next let x0 2 .a; b/ with f .x0�/ < f .x0C/. By the density of rationalnumbers, we may choose a q.x0/ 2 Q between f .x0�/ and f .x0C/. Thefunction q: J ! Q is injective because if f is discontinuous at x0, x1 withx0 < x1, then

q.x0/ < f .x0C/ � f�x0 C x1

2

� � f .x1�/ < q.x1/:By the injection theorem, J is countable, i.e. f has countably many discontinuouspoints on .a; b/: The cases of decreasing functions or other kinds of intervals aresimilar.

Appendix: Infinite Limits and Limit at Infinity

We begin with a definition of a sequence havingC1 as limit. In this case, the sequence does not have any upperbound in R; i.e. the sequence will pass any fixed r 2 R eventually and keep on going. Sequences with �1 as limitand functions with�1 limit are defined similarly.

34

Definitions. (1) A sequence fxng diverges toC1 (or has limitC1) iff for every r 2 R; there is K 2 N (depending onr) such that n � K implies xn > r: Similarly, a sequence fxng diverges to �1 (or has limit�1) iff for every r 2 R;there is K 2 N (depending on r) such that n � K implies xn < r:

(2) A function f : S ! Rdiverges to C1 (or has limit C1) as x tends to x0 in S iff for every r 2 R; there is� > 0 (depending on r and x0) such that for every x 2 S; 0 < jx � x0j < � implies f .x/ > r: Similarly, a functionf : S ! Rdiverges to�1 (or has limit�1) as x tends to x0 in S iff for every r 2 R; there is � > 0 (depending on rand x0) such that for every x 2 S; 0 < jx � x0j < � implies f .x/ < r:Exercise. Prove that if fxng is an increasing sequence, then either the limit of fxng exists (as a real number) or the limitis C1: (The decreasing case is similar withC1 replaced by �1:)

Limit at infinity for functions are defined similarly as for sequences as follow.

Definitions. (1) Let f : S ! Rbe a function such that C1 is an accumulation point of S (i.e. there is a sequencein S diverges to C1:) We say f .x/ converges to L (or has limit L) as x tends to C1 in S iff for every " > 0; thereis K 2 R such that for every x 2 S; x � K implies j f .x/ � L j < ": Similarly, let f : S ! R be a function suchthat �1 is an accumulation point of S (i.e. there is a sequence in S diverges to �1:) We say f .x/ converges to L(or has limit L) as x tends to �1 in S iff for every " > 0; there is K 2 R such that for every x 2 S; x � K impliesj f .x/ � L j < ":

(2) Let f : S ! R be a function such that C1 is an accumulation point of S (i.e. there is a sequence in Sdiverges to C1:) We say f .x/ diverges to C1 (or has limitC1) as x tends to C1 in S iff for every r 2 R; there isK 2 R such that for every x 2 S; x � K implies f .x/ > r: Similarly, let f : S ! Rbe a function such that �1 isan accumulation point of S (i.e. there is a sequence in S diverges to �1:) We say f .x/ diverges to C1 (or has limitC1) as x tends to �1 in S iff for every r 2 R; there is K 2 R such that for every x 2 S; x � K implies f .x/ > r:By replacing f .x/ > r with f .x/ < r; we get the definitions of limit equal to �1 as x tends to�1 in S:

It is good exercises for the readers to formulate the computation formulas and properties for these limits, whichcan be proved by checking these definitions just like the finite cases.

35

Chapter 7. Continuity

Definitions. A function f : S ! R is continuous at x0 2 S iff limx!x0x2S

f .x/ D f .x0/; i.e. for every " > 0, there is � > 0

such that for all x 2 S, jx � x0j < � ) j f .x/ � f .x0/j < ". For E � S; we say f is continuous on E iff f iscontinuous at every element of E : Also, we say f is continuous iff f is continuous on the domain S:Sequential Continuity Theorem. f : S ! R is continuous at x0 2 S if and only if for every xn ! x0 in S;lim

n!1 f .xn/ D f .x0/ D f . limn!1 xn/:

Proof. Just replace L by f .x0/, 0 < jx � x0j < � by jx � x0j < � and xn ! x0 in S n fx0g by xn ! x0 in S (i.e. deletethe xn 6D x0 requirement) in the proof of the sequential limit theorem.

Example. It is easy to give examples of continuous functions, such as polynomials. Here is an example of a function

not continuous at any point. Let f .x/ D � 1 if x 2 Q0 if x 62 Q ; then f is discontinuous at every x 2 R!

Reason. For every x0 2 R; n 2 N, by the density of rational numbers and irrational numbers, there are rn 2 Q, sn 62 Qin .x0 � 1

n ; x0/: Now rn ! x0, sn ! x0, but limn!1 f .rn / D 1 and lim

n!1 f .sn/ D 0. So limx!x0

f .x/ cannot exist.

Theorem. If f; g : S ! Rare continuous at x0 2 S; then f � g; f g; f =g (provided g.x0/ 6D 0) are continuous at x0:Proof. Since f; g are continuous at x0; lim

x!x0. f C g/.x/ D lim

x!x0f .x/ C lim

x!x0g.x/ D f .x0/C g.x0/ D . f C g/.x0/:

So f C g is continous at x0 by definition. The subtraction, multiplication and division cases are similar.

Theorem. If f : S ! R is continuous at x0, f .S/ � S0; g : S0 ! R is continuous at f .x0/, then g B f is continuousat x0.

Proof. By the sequential continuity theorem, all we need to show is that limn!1.g � f /.xn/ D .g � f /.x0/ for every

sequence fxng converging to x0: Since f is continuous at x0, by the sequential continuity theorem, the limit of f .xn/is f .x0/: Since g is continuous at f .x0/; so lim

n!1 g. f .xn// D g. limn!1 f .xn// D g. f .x0//:

In the discussions below, S will denote an interval of positive length.

Theorem (Sign Preserving Property). If g: S ! R is continuous and g.x0/ > 0, then there is an interval I D.x0� �; x0C �/ with � > 0 such that g.x/ > 0 for every x 2 S \ I: (The case g.x0/ < 0 is similar by considering�g.)

Proof. Let " D g.x0/ > 0: Since g is continuous at x0; there is � > 0 such that for x 2 S; jx � x0j < � )jg.x/ � g.x0/j < ": So x 2 S \ .x0 � �; x0 C �/) 0 D g.x0/ � " < g.x/:Intermediate Value Theorem. If f : [a; b]! R is continuous and y0 is between f .a/ and f .b/ inclusive, then thereis (at least one) x0 2 [a; b] such that f .x0/ D y0.

Proof. The cases y0 D f .a/ or y0 D f .b/ are trivial as x0 D a or b will do. We assume f .a/ < y0 < f .b/. (Theother possibility f .a/ > y0 > f .b/ is similar.) The set S D fx 2 [a; b]: f .x/ � y0g is nonempty (a 2 S) and has b asan upper bound. Then x0 D sup S 2 [a; b].

By the supremum limit theorem, there is a sequence fxng in S such that limn!1 xn D x0. Note xn 2 [a; b] implies

x0 2 [a; b]: Then f .x0/ D limn!1 f .xn/ � y0 by continuity of f at x0. Assume f .x0/ < y0. Then x0 6D b as y0 < f .b/:

Define g.x/ D y0 � f .x/ on [a; b]. Since g.x0/ > 0, by the sign preserving property, there is x1 > x0 such thatg.x1/ > 0: Then f .x1/ < y0. So x0 < x1 2 S; which contradicts x0 D sup S. Therefore, f .x0/ D y0.

Examples. (1) The equation x5 C 3x C sin x D cos x C 10 has a solution. To see this, let f .x/ D x5 C 3x C sin x�cos x�10; then f .0/ D �11 and 26 � f .2/ � 30: Since f is continuous, the intermediate value theorem impliesf .x/ D 0 for some x 2 .0; 2/:

36

(2) Suppose p.x/ D xn C a1xn�1 C � � � C an with n odd. Let x0 D 1 C ja1j C � � � C janj � 1; then we havep.x0/ D xn

0 C a1xn�10 C � � � C an and p.�x0/ D �xn

0 C a1xn�10 � � � �C an: So xn

0 � p.x0/ D �a1xn�10 � � � �� an

and xn0 C p.�x0/ D a1xn�1

0 � � � � C an : By the triangle inequality,

xn0 � p.x0/

xn0 C p.�x0/ � � ja1xn�1

0 j C � � � C janj � ja1jxn�10 C � � � C janjxn�1

0 D .ja1j C � � � C janj/xn�10 < xn

0 :Then p.x0/ > 0 and p.�x0/ < 0: By the intermediate value theorem, there is a root of p.x/ between�x0 and x0:

a x w b0 0

Extreme Value Theorem. If f : [a; b] ! R is continuous, then there are x0,w0 2 [a; b] such that f .w0/ � f .x/ � f .x0/ for every x 2 [a; b]: So the rangeof f is f

�[a; b]

� D [ f .w0/; f .x0/]: In particular, f is bounded on [a; b]

In this case, we write f .x0/ D supf f .x/ : x 2 [a; b]g D maxx2[a;b]

f .x/ and

f .w0/ D inff f .x/ : x 2 [a; b]g D minx2[a;b]

f .x/:Proof. We first show f .[a; b]/ D f f .x/ : x 2 [a; b]g is bounded above. Assume it is not bounded above. Then eachn 2 N is not an upper bound. So there is zn 2 [a; b] such that f .zn/ > n: By the Bolzano-Weierstrass theorem, fzngin [a; b] has a subsequence fznj g converging to some z0 2 [a; b]: Since f is continuous at z0; lim

j!1 f .znj / D f .z0/;which implies f f .znj /g is bounded by the boundedness theorem. However, f .znj / > n j � j implies f f .znj /g is notbounded, a contradiction.

By the completeness axiom, M D supf f .x/ : x 2 [a; b]g exists. By the supremum limit theorem, there is asequence fxng in [a; b] such that lim

n!1 f .xn/ D M . By the Bolzano-Weierstrass theorem, fxng has a subsequence fxnk gsuch that lim

k!1 xnk D x0 in [a; b]. Since f is continuous at x0 2 [a; b], M D limk!1 f .xnk / D f .x0/ by the subsequence

theorem and the sequential continuity theorem, respectively.

Similarly, inff f .x/ : x 2 [a; b]g D f .w0/ for some w0 2 [a; b].

The following two theorems are for explaining thedx

dyD 1

�dy

dxrule in the next chapter.

Continuous Injection Theorem. If f is continuous and injective on [a; b], then f is strictly monotone on [a; b] andf�[a; b]

� D [ f .a/; f .b/] or [ f .b/; f .a/]. (This is true for any nonempty interval in place of [a; b]: The range is aninterval with f .a/; f .b/ as endpoints.)

f (y)

f (b)

f (a)

a y b

Proof. Since f is injective, either f .a/ < f .b/ or f .a/ > f .b/. Supposef .a/ < f .b/. Let y 2 .a; b/. Then f .y/ cannot be greater than f .b/, otherwiseby the intermediate value theorem, there is w 2 .a; y/ such that f .w/ D f .b/,contradicting injectivity. Similarly, f .y/ cannot be less than f .a/. Therefore,f .a/ < f .y/ < f .b/. If a � x < y � b, then similarly f .a/ � f .x/ <f .y/ � f .b/, i.e. f is strictly increasing on [a; b] and f

�[a; b]

�D [ f .a/; f .b/]by the intermediate value theorem.

The case f .a/ > f .b/ leads to f strictly decreasing on [a; b].

Continuous Inverse Theorem. If f is continuous and injective on [a; b], then f �1 is continuous on f�[a; b]

�.

Proof. By the last theorem, f is strictly monotone on [a; b]. We first suppose f is strictly increasing. Then f �1 isalso strictly increasing on f

�[a; b]

� D [ f .a/; f .b/].For y0 2 . f .a/; f .b/], let r D f �1.y0�/ D lim

y!y�0 f �1.y/; which exists by the monotone function theorem.

Since f �1.y/ 2 [a; b]; r 2 [a; b]: Let fyng be a sequence in . f .a/; y0/ converging to y0; then wn D f �1.yn/will converge to r by sequential limit theorem. Since f is continuous at r; by the sequential continuity theorem,f .r/ D lim

n!1 f .wn/ D limn!1 f . f �1.yn// D y0: Then f �1.y0/ D r D f �1.y0�/. Similarly, if y0 2 [ f .a/; f .b//, then

f �1.y0C/ D f �1.y0/. So f �1 is continuous on [ f .a/; f .b/] D f�[a; b]

�. The case f is strictly decreasing is similar.

37

Appendix: Fundamental Theorem of Algebra

The extreme value theorem is often used in estimating integrals. More precisely, if f : [a; b] ! Ris continuous,

then for m D minx2[a;b]

f .x/ and M D maxx2[a;b]

f .x/; we have m � f .x/ � M and m.b � a/ � Z b

af .x/ dx � M.b � a/:

Here we will mention that there is a version of the extreme value theorem for continuous functions defined onclosed disks of finite radius on the plane. As an application of this fact, we can sketch a proof of the fundamentaltheorem of algebra, which asserts that every nonconstant polynomial with complex coefficients must have a root.

Let p.z/ D zn C a1zn�1 C � � � C an and m D inffjp.z/j : z 2 C g: We havejznj D ���p.z/ � nXkD1

akzn�k��� � jp.z/j C nX

kD1

jak jjzjn�k ) jp.z/j � jzjn � nXkD1

jakjjzjn�k D jzjn�1� nXkD1

jak jjzj�k| {z }!1� as jzj!1 �:So for jzj large, 1� nX

kD1

jak jjzj�k � 1

2: For a very large K > n

p2m; we have jzj > K impliesjp.z/j � jzjn�1� nX

kD1

jak jjzj�k� � 1

2jzjn > 1

2K n > m:

Let DK be the closed disk fz : jzj � K g: We have m D inffjp.z/j : z 2 C g D inffjp.z/j : z 2 DK g D jp.z0/j for somez0 2 DK by the extreme value theorem, since jp.z/j is continuous on DK :

We claim m D jp.z0/j D 0: Suppose m 6D 0: Then f .z/ D p.z C z0/=p.z0/ is a polynomial of degree n andf .0/ D 1: So f .z/ D 1Cb1zC� � �Cbnzn for some b1; : : : ; bn with bn 6D 0: Let k be the smallest positive integer suchthat bk 6D 0: Then f .z/ D 1C bkzk C � � � C bnzn : Since jp.z C z0/j � m D jp.z0/j; we get .�/ j f .z/j � 1 for all z:

Introduce the notation ei� D cos�C i sin�;which describes the points on the unit circle. Since the absolute valueof �jbkj=bk is 1, so �jbk j=bk D ei� for some �: Let � D �=k; then eik�bk D ei�bk D �jbkj: Considering w D rei�with r < jbkj�1=k.) 1� rk jbkj > 0/; we have j1C bkwkj D j1C bkrk eik� j D ��1� rk jbk j�� D 1� rk jbk j andj f .w/j D j1C bkwk C � � � C bnwnj � j1C bkwk j C jbkC1wkC1j C � � � C jbnwnjD 1� rk jbk j C jbkC1jrkC1 C � � � C jbnjrnD 1� rk�jbkj � jbkC1jr � � � � � jbnjrn�k| {z }!jbk j> 1

2 jbk j>0 as r!0C �:Taking w D rei� with a very small positive r; we have j f .w/j � 1� rk . 1

2 jbkj/ < 1; a contradiction to .�/: Thereforem D jp.z0/j D 0 and z0 is a root of p.z/:

38

Chapter 8. Differentiation

Definitions. Let S be an interval of positive length. A function f : S ! R is differentiable at x0 2 S iff f 0.x0/ Dlim

x!x0x2S

f .x/ � f .x0/x � x0

exists inR. Also, f is differentiable iff f is differentiable at every element of S:Theorem. If f : S ! R is differentiable at x0, then f is continuous at x0.

Proof. By the computation formulas for limits,

limx!x0

f .x/ D limx!x0

��f .x/ � f .x0/

x � x0

�.x � x0/C f .x0/� D f 0.x0/ � 0C f .x0/ D f .x0/:Theorem (Differentiation Formulas). If f; g : S ! Rare differentiable at x0, then f C g, f � g, f g, f=g (wheng.x0/ 6D 0) are differentiable at x0. In fact, . f � g/0.x0/ D f 0.x0/ � g0.x0/, . f g/0.x0/ D f 0.x0/g.x0/ C f .x0/g0.x0/and . f

g/0.x0/ D f 0.x0/g.x0/ � f .x0/g0.x0/

g.x0/2.

Proof. By the computation formulas for limits,

limx!x0

. f � g/.x/ � . f � g/.x0/x � x0

D limx!x0

�f .x/ � f .x0/

x � x0� g.x/� g.x0/

x � x0

� D f 0.x0/� g0.x0/;lim

x!x0

. f g/.x/ � . f g/.x0/x � x0

D limx!x0

�f .x/ � f .x0/

x � x0g.x/ C f .x0/g.x/ � g.x0/

x � x0

� D f 0.x0/g.x0/C f .x0/g0.x0/;lim

x!x0

. f

g/.x/ � . f

g/.x0/

x � x0D lim

x!x0

1g.x/g.x0/ � f .x/ � f .x0/

x � x0g.x0/ � g.x/ � g.x0/

x � x0f .x0/�D f 0.x0/g.x0/� f .x0/g0.x0/

g.x0/2:

Theorem (Chain Rule). If f : S ! R is differentiable at x0; f .S/ � S0 and g : S0 ! R is differentiable at f .x0/,then g B f is differentiable at x0 and .g B f /0.x0/ D g0. f .x0// f 0.x0/.Proof. The function

h.y/ D 8<: g.y/ � g. f .x0//y � f .x0/ if y 6D f .x0/

g0. f .x0// if y D f .x0/is continuous at f .x0/ because lim

y! f .x0/ h.y/ D g0. f .x0// D h. f .x0//. So g.y/ � g. f .x0// D h.y/.y � f .x0// holds

for every y in the domain of g. Then

limx!x0

g B f .x/ � g B f .x0/x � x0

D limx!x0

h. f .x// f .x/ � f .x0/x � x0

D h. f .x0// f 0.x0/ D g0. f .x0// f 0.x0/:Remarks. Note f differentiable at x0 does not imply f 0 is continuous at x0: In fact, the function f .x/ D x2 sin 1

x for

x 6D 0 and f .0/ D limx!0

x2 sin1xD 0 is continuous and differentiable with f 0.x/ D 2x sin 1

x � cos 1x for x 6D 0 and

f 0.0/ D limx!0

.x2 sin1

x�0/=x D 0: However, f 0 is not continuous at 0 because lim

x!0f 0.x/ D lim

x!02x sin

1

x� cos

1

xdoes

not exist, hence not equal to f 0.0/: In particular, f is not twice differentiable. (Also, the function g.x/ D x2 sin1

x2

for x 6D 0 and g.0/ D 0 is differentiable onR; but g0.x/ is not continuous at 0 and g0.x/ is unbounded on every openinterval containing 0.)

39

Notations. C0.S/ D C.S/ is the set of all continuous functions on S: For n 2 N;Cn.S/ is the set of all functionsf : S ! Rsuch that the n-th derivative f .n/ is continuous on S. C1.S/ is the set of all functions having n-th derivativesfor all n 2 N. Functions in C1.S/ are said to be continuously differentiable on S:Inverse Function Theorem. If f is continuous and injective on .a; b/ and f 0.x0/ 6D 0 for some x0 2 .a; b/, then f �1

is differentiable at y0 D f .x0/ and . f �1/0.y0/ D 1

f 0.x0/ , i.e.dx

dyD 1

�dy

dx.

Proof. The function g.x/ D 8><>: x � x0

f .x/ � f .x0/ if x 6D x0

1

f 0.x0/ if x D x0

is continuous at x0 because limx!x0

g.x/ D 1

f 0.x0/ D g.x0/:Since f is continuous and injective on .a; b/; f �1 is continuous by the continuous inverse theorem. So lim

y!y0f �1.y/ D

f �1.y0/ D x0 and . f �1/0.y0/ D limy!y0

f �1.y/ � f �1.y0/y � y0

D limy!y0

g�

f �1.y/� D g.x0/ D 1

f 0.x0/ :Local Extremum Theorem. Let f : .a; b/ ! Rbe differentiable. If f .x0/ D min

x2.a;b/ f .x/ or f .x0/ D maxx2.a;b/ f .x/;

then f 0.x0/ D 0:Proof. If f .x0/ D min

x2.a;b/ f .x/; then 0 � limx!xC0 f .x/ � f .x0/

x � x0D f 0.x0/ D lim

x!x�0 f .x/ � f .x0/x � x0

� 0: So f 0.x0/ D 0:The other case is similar.

w0

x0

Rolle’s Theorem. Let f be continuous on [a; b] and differentiable on .a; b/. Iff .a/ D f .b/; then there is (at least one) z0 2 .a; b/ such that f 0.z0/ D 0.

Proof. This is trivial if f is constant on [a; b]. Otherwise, by the extremevalue theorem, there are x0, w0 2 [a; b] such that f .x0/ D max

x2[a;b]f .x/ >

minx2[a;b]

f .x/ D f .w0/. Either f .x0/ 6D f .a/ or f .w0/ 6D f .a/. So either x0 orw0 is in .a; b/: By the last theorem, f 0.x0/ D 0 or f 0.w0/ D 0:a x0 b

Mean-Value Theorem. Let f be continuous on [a; b] and differentiable on.a; b/. Then there exists x0 2 .a; b/ such that f .b/� f .a/ D f 0.x0/.b � a/.Proof. Define F.x/ D f .x/ � � f .b/ � f .a/

b � a.x � a/C f .a/�. Then F.a/ D

0 D F.b/. By Rolle’s theorem, there is x0 2 .a; b/ such that 0 D F 0.x0/ Df 0.x0/ � f .b/ � f .a/

b � a, which is equivalent to f .b/ � f .a/ D f 0.x0/.b � a/:

Examples. (1) For a < b; show j sin b � sin aj � jb � aj: (By the mean-value theorem, there is x0 2 .a; b/ such thatsin b � sin a D .cos x0/.b � a/; so j sin b � sin aj � jb � aj:)

(2) Show .1 C x/� � 1 C �x for x � �1 and � � 1: (Let f .x/ D .1 C x/� � 1� �x; then there is x0 2 .0; x/ ifx > 0 or x0 2 .x; 0/ if �1 < x < 0 such that.1C x/� � 1� �x D f .x/ � f .0/ D f 0.x0/.x � 0/ D �..1 C x0/��1 � 1/x � 0:/

(3) Show ln x � x � 1 for x > 0: (Let f .x/ D ln x � x C 1; then f .1/ D 0: If x > 1; then there is x0 2 .1; x/ suchthat

ln x � x C 1 D f .x/ D f .x/ � f .1/ D f 0.x0/.x � 1/ D � 1x0� 1

� .x � 1/ � 0:The case 0 < x < 1 is similar.)

40

(4) To approximatep

16:1; we can let f .x/ D px : Then f .16:1/ � f .16/ D f 0.c/.16:1 � 16/ for some c

between 16 and 16:1. Now c � 16 and f .16:1/ � f .16/ � f 0.16/.16:1 � 16/ D 0:0125; which givesp16:1 D f .16:1/ � 4:0125:

Curve Tracing Theorem. If f 0 � 0 (respectively f 0 > 0; f 0 � 0; f 0 < 0; f 0 6D 0; f 0 � 0) everywhere on .a; b/, thenf is increasing (respectively strictly increasing, decreasing, strictly decreasing, injective, constant) on .a; b/:Proof. If x , y 2 .a; b/ and x < y, then by the mean value theorem, there is x0 2 .x; y/ such that f .y/ �f .x/ D f 0.x0/.y � x/ � 0 (respectively > 0; � 0; < 0; 6D 0; D 0). Therefore, f .x/ � f .y/ (respectivelyf .x/ < f .y/; f .x/ � f .y/; f .x/ > f .y/; f .x/ 6D f .y/; f .x/ D f .y/).Remarks. For differentiable function f; the converse of the strictly increasing (respectively strictly decreasing,injective) parts of the curve tracing theorem are false. To see an example, consider f .x/ D x3; which is strictlyincreasing on R; but f 0.0/ D 0: The converse of the increasing (respectively decreasing, constant) parts are truebecause . f .x/� f .x0//=.x � x0/ is nonnegative (respectively nonpositive, zero) for x; x0 2 .a; b/ and hence the samefor the limit as x tends to x0 on .a; b/:Local Tracing Theorem. If f : [a; b] ! Ris continuous and f 0.c/ > 0 for some c 2 [a; b]; then there are c0; c1 2 Rsuch that c0 < c < c1 and f .x/ < f .c/ < f .y/ for all x 2 .c0; c/\ [a; b] and all y 2 .c; c1/\ [a; b]:A similar resultfor the case f 0.c/ < 0 holds and the inequality becomes f .x/ > f .c/ > f .y/:Proof. Let f 0.c/ > 0: Assume there is no such c0: Then for every n 2 N; there is xn 2 .c � 1

n ; c/ \ [a; b] such that

f .xn/ � f .c/: This leads to f 0.c/ D limn!1 f .xn/ � f .c/

xn � c� 0; a contradiction. The other parts of the theorem are

similar.

Remarks. If f 0.c/ � 0; we may not have f .x/ � f .c/ � f .y/ similar to above as the function f .x/ D x2 sin1x

with

f .0/ D 0 satisfies f 0.0/ D 0; but f takes positive and negative values on every open interval about 0.

Exercise. A function f : [a; b] ! Ris said to have the intermediate value property iff for every y0 in the open intervalwith endpoints f .a/ and f .b/; there exists at least one x0 2 .a; b/ such that f .x0/ D y0: We have already showedthat continuous functions on [a; b] satisfied this property. Prove that if g is differentiable on [a; b]; then g0 has theintermediate value property. In particular, if g0.x/ 6D 0 for all x 2 [a; b]; then g0 > 0 or g0 < 0 everywhere on [a; b]:

Next we will introduce the generalized mean-value theorem, which has two very important applications, namelyTaylor’s theorem and L’Hopital’s rule.

Generalized Mean-Value Theorem. Let f , g be continuous on [a; b] and differentiable on .a; b/. Then there isx0 2 .a; b/ such that g0.x0/. f .b/� f .a// D f 0.x0/.g.b/�g.a//. (Note the case g.x/ D x is the mean-value theorem.)

Proof. Let F.x/ D f .x/.g.b/ � g.a//� . f .b/� f .a//.g.x/ � g.a//, then F.a/ D f .a/.g.b/� g.a// D F.b/. ByRolle’s theorem, there is x0 2 .a; b/ such that 0 D F 0.x0/ D f 0.x0/.g.b/ � g.a//� g0.x0/. f .b/ � f .a//:Taylor’s Theorem. Let f : .a; b/! Rbe n times differentiable on .a; b/. For every x, c 2 .a; b/, there is x0 betweenx and c such that

f .x/ D f .c/ C f 0.c/1!

.x � c/ C f 00.c/2!

.x � c/2 C : : : C f .n�1/.c/.n � 1/! .x � c/n�1 C f .n/.x0/n!

.x � c/n :(This is called the n-th Taylor expansion of f about c: The term Rn.x/ D f .n/.x0/

n!.x � c/n is called the Lagrange

form of the remainder.)

Proof. Let I be the closed interval with endpoints x and c. For t 2 I , define g.t/ D .n � 1/!n�1XkD0

f .k/.t/k!

.x � t/k;where f .0/ D f; .x � x/0 D 1 and define p.t/ D � .x � t/n

n: We have g0.t/ D f .n/.t/.x � t/n�1 and p0.t/ D

41

.x� t/n�1 :By the generalized mean value theorem, there is x0 between x and c such that g0.x0/| {z }f .n/.x0/.x�x0/n�1

[p.x/ � p.c/| {z }.x�c/n=n

] Dp0.x0/| {z }.x�x0/n�1

[ g.x/|{z}.n�1/! f .x/�g.c/]. Then f .x/ D g.c/.n � 1/! C f .n/.x0/n!

.x � c/n D n�1XkD0

f .k/.c/k!

.x � c/k C f .n/.x0/n!

.x � c/n :Lemma. Let h : .a; b/ ! [0;C1/ be a bounded function, where a may be real or �1 and b may be real orC1: We have lim

x!aC h.x/ D 0 if and only if limx!aC supfh.t/ : a < t < xg D 0: Similarly, lim

x!b� h.x/ D 0 if and only if

limx!b�supfh.t/ : x < t < bgD 0:Proof. For right hand limits at a; lim

x!aC h.x/ D 0 is the same as for every " > 0; there exists r 2 R such that

for all a < x < r implies h.x/ D jh.x/ � 0j < ": The later part is the same as for all a < x < r impliessupfh.t/ : a < t < xg � "; which is lim

x!aC supfh.t/ : a < t < xg D 0: The statement for the left hand limits at b is

similar.

L’Hopital’s Rule ( 00 Version). Let f , g be differentiable on .a; b/ and g.x/; g0.x/ 6D 0 for all x 2 .a; b/; where�1 � a < b � C1: If

(a) limx!aC f 0.x/

g0.x/ D L ; where �1 � L � C1; and

(b) limx!aC f .x/ D 0 D lim

x!aC g.x/;then lim

x!aC f .x/g.x/ D L : (Similarly, the rule is also true if x ! b�:)

Proof. For f 0=g0 ! L 2 R; replacing f=g by . f � Lg/=g D . f=g/ � L ; we may assume L D 0:Since lim

x!aC f 0.x/g0.x/ D 0; we may assume

f 0.x/g0.x/ is bounded on .a; c/ � .a; b/: For every x; y 2 .a; c/ with y < x;

by generalized mean-value theorem, there is w 2 .y; x/ such that��� f .x/ � f .y/g.x/ � g.y/ ��� D ��� f 0.w/

g0.w/ ��� � supn��� f 0.t/

g0.t/ ��� : a < t < xo: .�/

Taking limit of y ! aC on both sides, we get��� f .x/

g.x/ ��� � supn��� f 0.t/

g0.t/ ��� : a < t < xo: By condition (a) and the lemma,

the right side goes to 0. So by the sandwich theorem, we get the left side go to 0, which is the conclusion.

For f 0=g0 ! L D C1; there is r 2 R such that a < t < r implies f 0.t/=g0.t/ > 1 so that f 0 and g0 are both

positive or both negative when x 2 .a; r/: Next for a < y < x < r; we havef .x/ � f .y/g.x/� g.y/ D f 0.t/

g0.t/ > 1: As y ! aC;we see

f .x/g.x/ � 1: So f and g are both positive or both negative when x 2 .a; r/: Now g0= f 0 ! 0; g ! 0 and f ! 0

as x ! aC: So by above, g=f ! 0: Hence f=g !C1: The cases L D �1 and x ! b� are similar.

Example. (Even if limx!a

f .x/ D 0 D limx!a

g.x/ and limx!a

f 0.x/g0.x/ does not exist, lim

x!a

f .x/g.x/ may still exist as the following

example will show.) Let f .x/ D x2 sin1x

, g.x/ D sin x , a D 0, then �x2 � x2 sin1x� x2 implies lim

x!0f .x/ D 0

by sandwich theorem, limx!0

g.x/ D 0, limx!0

f 0.x/g0.x/ D lim

x!0

2x sin 1x � cos 1

x

cos xD lim

x!0

0� cos 1x

1does not exist. However,

limx!0

f .x/g.x/ D lim

x!0

� x

sin x

�.x sin1x/ D 1 � 0 D 0.

L’Hopital’s Rule ( �1 Version). Let f , g be differentiable on .a; b/ and g.x/; g0.x/ 6D 0 for all x 2 .a; b/; where�1 � a < b � C1: If

42

(a) limx!aC f 0.x/

g0.x/ D L ; where �1 � L � C1; and

(b) limx!aC g.x/ D C1

then limx!aC f .x/

g.x/ D L : (Similarly, the rule is also true if x ! b� or g.x/!�1).

Proof. We will modify the proof above as follow. If L 2 R; then we can reduce to L D 0 as above. In the casef 0=g0 ! L D C1; for x; s 2 .a; b/; by generalized mean-value theorem, there is t between x and s such that

f .x/ D f .s/ C f 0.t/g0.t/ .g.x/ � g.s// !C1 as x ! aC by (a) and (b). So we may replace f=g by g= f to assume

L D 0: Multiplying the left and right sides of (*) by��� g.x/� g.y/

g.y/ ���; we get for a < y < x < c;��� f .x/ � f .y/g.y/ ��� � sup

n��� f 0.t/g0.t/ ��� : a < t < x

o| {z }! LD0 as x! aC by lemma

� ���g.x/ � g.y/g.y/ ���| {z }! 1 as y! aC :

For every " > 0; there is x 2 .a; b/ such that supn��� f 0.t/

g0.t/ ��� : a < t < xo < "

4: Also, lim

y!aC f .x/g.y/ D 0: Then there is

r < x such that for all y 2 .a; r/; we can get��� f .x/

g.y/ ���< "2

and��� g.x/� g.y/

g.y/ ��� < 2. Then��� f .y/g.y/ ��� � ��� f .x/

g.y/ ��� C ��� f .y/ � f .x/g.y/ ��� < "

2C "

4� 2 D ":

The conclusion follows. The cases L D �1; x ! b� and g.x/!�1 are similar.

Examples. (1) (Even if limx!a

f .x/ D C1 D limx!a

g.x/ and limx!a

f 0.x/g0.x/ does not exist, lim

x!a

f .x/g.x/ may still exist as the fol-

lowing example will show.) Let f .x/ D 2xCsin x , g.x/ D 2x�sin x , a D C1, then limx!C1 f .x/ D C1 D lim

x!C1 g.x/,lim

x!C1 f 0.x/g0.x/ D lim

x!C1 2C cos x

2� cos xdoes not exist. However, lim

x!C1 f .x/g.x/ D lim

x!C1 2C .sin x/=x

2� .sin x/=xD 1:

(2) Show that limx!C1 xr

exD 0 for every r 2 R:

Solution. There is an integer n � jrj (by the Archimedian principle). We have xr � xn on [1;C1/: So 0 � xr

ex� xn

ex

on [1;1/: Sincedn

dxnxn D n! and lim

x!C1 n!exD 0; applying L’Hopital’s rule n times, we see lim

x!C1 xn

exD 0; which

implies limx!C1 xr

exD 0 by the sandwich theorem.

(3) Let f : .a;C1/ ! R be differentiable. Show that limx!C1 f 0.x/ C f .x/ D 0 if and only if lim

x!C1 f .x/ D 0 and

limx!C1 f 0.x/ D 0: (This type of result is often needed in the studies of differential equations. Here if lim

x!C1 g.x/ D 0;then every solution y D f .x/ of the equation

dy

dxC y D g.x/ will have limit 0 as x !C1:)

Solution. The if part follows by the computation formulas for limit. For the only if part, consider f .x/ D f .x/ex

ex:Note

the denominator on the right side tend to C1 as x tend to C1: Since limx!C1 �

f .x/ex�0.ex /0 D lim

x!C1 f 0.x/ C f .x/ D 0;by L’Hopital’s rule, we get lim

x!C1 f .x/ D limx!C1 f .x/ex

exD 0 and lim

x!C1 f 0.x/ D limx!C1� f 0.x/ C f .x/� � f .x/ D 0:

43

Appendix 1: Convex and Concave Functions.

tf (a) 1(+ - t ) f (b)

ta+(1- t ) b ba

f (b)

f(a)

convex function

y = f (x)

Definitions. Let I be an interval and f : I ! R. We say f is a convexfunction on I iff for every a, b 2 I , 0 � t � 1,

f�ta C .1 � t/b� � t f .a/ C .1� t/ f .b/:

We say f is a concave function on I if

f�ta C .1 � t/b� � t f .a/ C .1� t/ f .b/:

Remarks. As t ranges from 0 to 1, the point�ta C .1 � t/b; f .ta C.1� t/b/� traces the graph of y D f .x/ for a � x � b, while the point�

taC .1� t/b; t f .a/C .1� t/ f .b/� traces the chord joining .a; f .a//to .b; f .b//. So f is convex on I if and only if the chord joining twopoints on the graph always lies above or on the graph.

(x, f (x))

(a, f ))a(

(b, f (b))

a x b

Theorem. f is convex on I if and only if the slope of the chords always

increase, i.e. a, x, b 2 I , a < x < b ) f .x/ � f .a/x � a

� f .b/ � f .x/b � x

.

Proof. x D taC .1� t/b for some t 2 [0; 1] () 0 � t D b� x

b � a� 1.

f .x/ � f .a/x � a

� f .b/ � f .x/b � x

() f .x/ � b � x

b � af .a/C x � a

b � af .b/ () f

�taC.1�t/b�� t f .a/C.1�t/ f .b/:

Theorem. For f differentiable on I; f is convex on I () f 0 is increasing on I . (For f twice differentiable on I;f is convex on I () f 00 � 0 on I .)

Proof. ()) If a, b 2 I , a < b, then f 0.a/ D limx!aC f .x/ � f .a/

x � a� lim

x!aC f .b/ � f .x/b � x

D f .b/ � f .a/b � a

Dlim

x!b� f .x/ � f .a/x � a

� limx!b� f .b/ � f .x/

b � xD f 0.b/.

(() If a, x , b 2 I , a < x < b, then by the mean value theorem, there are r, s such that a < r < x < s < b andf .x/ � f .a/

x � aD f 0.r/ � f 0.s/ D f .b/ � f .x/

b � x.

Theorem. If f is convex on .a; b/, then f is continuous on .a; b/.Proof. For x0 2 .a; b/, consider u, v, w 2 .a; b/ such that u < x0 < v < w. Then

f .x0/� f .u/x0 � u

� f .v/ � f .x0/v � x0�

f .w/ � f .v/w � v . Solving for f .v/; we getf .x0/ � f .u/

x0 � u.v � x0/ C f .x0/ � f .v/ � f .w/� f .v/w � v .v � x0/ C f .x0/:

Taking limit as v ! xC0 , we have f .x0/ � f .x0C/ � f .x0/, i.e. f .x0C/ D f .x0/. Similarly, we can showf .x0�/ D f .x0/ by taking u < v < x0 < w: Therefore, lim

x!x0f .x/ D f .x0/ for every x0 2 .a; b/.

Example. The function f .x/ D n0 if 0 � x < 11 if x D 1

is convex on [0; 1] by checking the slopes of the chords over [0; 1],

but f is not continuous at 1 because limx!1� f .x/ D 0 6D 1 D f .1/.

Remark. The proof of the last theorem uses the fact that on an open interval, at any point x0, there are points on itsleft and points on its right, which is not true for endpoints of a closed interval.

Example. Prove that if a; b � 0 and 0 < r < 1; then jar � br j � ja � bjr : (In particular, for n D 2; 3; 4; : : : ; wehave j n

pa � n

pbj � n

pja � bj:)Solution. We may assume a > b; otherwise interchange a and b:Define f : [0; a] ! Rby f .x/ D xrC.a�x/r : Sincer � 1 < 0; so f 00.x/ D r.r � 1/�xr�2 C .a � x/r�2

� � 0: So f is concave on [0; a]: Since f .0/ D ar D f .a/; we getf .x/ D xrC.a�x/r � ar for all x 2 [0; a]:Therefore, if b 2 [0; a];we have ja�bjr D .a�b/r � ar�br D jar�br j:

44

Chapter 9. Riemann Integral

For a; b 2 R; let f be a bounded function on [a; b]; say j f .x/j � K for every x 2 [a; b]. Let P D fx0; x1; : : : ; xngbe a partition of [a; b], i.e. a D x0 < x1 < : : : < xn�1 < xn D b. The length of [x j�1; x j] is 1x j D x j � x j�1

and the mesh of P is kPk D maxf1x1; : : : ; 1xng: Also, on [x j�1; x j]; let m j D inff f .x/ : x 2 [x j�1; x j]g andMj D supf f .x/ : x 2 [x j�1; x j]g:

a = x0 x1 x2 . . . xn - 1 xn = b

Definitions. For the partition P; a Riemann sum is a sum of the form S DnX

jD1

f .tj/1x j ;where every tj is in [x j�1; x j]: The lower Riemann sum is L. f; P/ DnX

jD1

m j1x j and the upper Riemann sum is U . f; P/ D nXjD1

Mj1x j :(Note �K � m j � f .tj / � Mj � K implies �K .b � a/ � L. f; P/ � S �U . f; P/ � K .b � a/: )

In 1854, George B. Riemann defined the integral of f .x/ on [a; b] to be limkPk!0

nXjD1

f .tj/1x j : There are two

immediate technical problems with such a definition.

(1) As the choices of the tj’s may vary, it is hard to say how close the Riemann sum is to the actual integral. Inparticular, it is not clear if the Riemann sum is greater than or less than the integral at any time.

(2) The type of limit involved is not the limit of a sequence nor the limit of a function. In fact, there are many variablesin a Riemann sum!

Instead of dealing with these technicalities, we introduce integral following the approach of J. Gaston Darboux in 1875.

Definition. We say P 0 is a refinement of P or P 0 is finer than P iff P and P 0 are partitions of [a; b] and P � P 0:Refinement Theorem. If P 0 is a finer partition of [a; b] than P, then L. f; P/ � L. f; P 0/ � U . f; P 0/ � U . f; P/.

x xw

w

jj - 1

j

Proof. P 0 can be obtained from P by adding one point at a time. It suffices to considerthe case P 0 D P [ fwg with w 62 P D fx0; x1; : : : ; xng. Suppose x j�1 < w < x j :

Let m j D inff f .x/ : x 2 [x j�1; x j]g; m0j D inff f .x/ : x 2 [x j�1; w]g and m00

j Dinff f .x/ : x 2 [w; x j]g: Since [x j�1; w]; [w; x j] � [x j�1; x j]; we have m j � m0

j andm j � m00

j : Then m j1x j D m j .w � x j�1/ Cm j .x j � w/ � m0j .w � x j�1/ C m00

j .x j � w/:So L. f; P/ � L. f; P 0/. Similarly, U . f; P 0/ � U . f; P/.

Observe that L. f; P/ “underestimates” the area under the curve; U . f; P/ “overestimates” the area under thecurve.

Definitions. The lower integral of f on [a; b] is.L/ Z b

af .x/ dx D supfL. f; P/: P is a partition of [a; b]g (the largest underestimate/

and the upper integral of f on [a; b] is.U / Z b

af .x/ dx D inffU . f; P/: P is a partition of [a; b]g (the smallest overestimate):

(So for every partition P of [a; b]; L. f; P/ � .L/ R ba f .x/ dx � .U / R b

a f .x/ dx � U . f; P/:)We say f is (Riemann) integrable on [a; b] iff .L/ R b

a f .x/ dx D .U / R ba f .x/ dx : In that case, we denote the

common value byR b

a f .x/ dx . (For b � a, defineR b

a f .x/ dx D � R ab f .x/ dx . In particular,

R aa f .x/ dx D 0.)

45

U( f, P) - L ( f, P)= sum of areas

of shaded regions

Theorem (Integral Criterion). Let f : [a; b] ! Rbe a bounded function. Thefunction f is (Riemann) integrable on [a; b] if and only if for every " > 0, thereis a partition P of [a; b] such that U . f; P/ � L. f; P/ < ":

Proof. If for every " > 0; there is a partition P of [a; b] such that U . f; P/� L. f; P/ < ", then 0 � .U / R ba f .x/ dx �.L/ R b

a f .x/ dx � U . f; P/ � L. f; P/ < ": By the infinitesimal principle, we get .L/ R ba f .x/ dx D .U / R b

a f .x/ dx;i.e. f is (Riemann) integrable on [a; b]:

Conversely, if f is (Riemann) integrable on [a; b]; then for every " > 0; by the supremum property, we have.L/ R ba f .x/ dx � "=2 < L. f; P1/ � .L/ R b

a f .x/ dx for some partition P1 of [a; b]: Similarly, .U / R ba f .x/ dx �

U . f; P2/ < .U / R ba f .x/ dx C "=2 for some partition P2 of [a; b]: Let P D P1 [ P2; then by the refinement

theorem, L. f; P1/ � L. f; P/ � U . f; P/ � U . f; P2/: Since U . f; P2/ � L. f; P1/ < [.U / R ba f .x/ dx C "=2] �

[.L/ R ba f .x/ dx � "=2] D "; so U . f; P/ � L. f; P/ � U . f; P2/ � L. f; P1/ < ":

Question. Are there integrable functions? Are there non-integrable functions?

Below we will show that constant functions and continuous functions are integrable.

Example. Recall the function f .x/ D �1 if x 2 Q0 if x 62 Q is not continuous anywhere. Partitioning an interval [a; b] into

subintervals [x j�1; x j], we get by the density of rational numbers and irrational numbers that m j D 0 and Mj D 1: Then

L. f; P/ D 0 and U . f; P/ D b� a for every partition P of [a; b]: So .L/ Z b

af .x/ dx D 0, .U / Z b

af .x/ dx D b� a.

Therefore, f .x/ is not integrable on every interval [a; b] with a < b.

Example. If f .x/ D c for every x 2 [a; b]; then L. f; P/ D c.b � a/ D U . f; P/ for every partition P of [a; b]: So.L/ Z b

af .x/ dx D c.b � a/ D .U / Z b

af .x/ dx : Therefore, f is integrable on [a; b] and

Z b

af .x/ dx D c.b � a/:

Uniform Continuity Theorem. If f : [a; b] ! R is continuous, then f is uniformly continuous (in the sense that forevery " > 0; there exists a � > 0 such that for all x; t 2 [a; b]; jx � t j < � implies j f .x/ � f .t/j < ":)Proof. Suppose f is not uniform continuous. Then there is an " > 0 such that for every � D 1

n ; n 2 N; there arexn; tn 2 [a; b] such that jxn � tnj < � D 1

n and j f .xn/ � f .tn/j � ": By the Bolzano-Weierstrass theorem, fxng has

a subsequence fxnj g converging to some w 2 [a; b]: Since jtnj � wj � jtnj � xnj j C jxnj � wj � 1

n jC jxnj � wj; by

the sandwich theorem, ftnj g converges to w: Then 0 D j f .w/� f .w/j D limj!1 j f .xnj /� f .tnj /j � " > 0; which is a

contradiction.

Examples. For � 2 .0; 1]; a function f : S ! R is Holder continuous of order � iff f staisfies the condition thatthere exists constant M > 0 such that for all x; y 2 S; j f .x/ � f .y/j � Mjx � yj�: In case � D 1; f is said tobe Lipschitz. The constant M is called the Lipschitz constant or Holder constant for f: All functions with boundedderivatives (such as cos x with M D 1) are Lipschitz by the mean-value theorem. The function f .x/ D p

x satisfiesjpx �pyj � pjx � yj on [0;C1/ is Holder continuous of order 1

2 : All Holder functions are uniformly continuous,since for every " > 0; we can take � D ."=m/1=�:Theorem. If f is continuous on [a; b], then f is integrable on [a; b].

Proof. For " > 0, since f is uniformly continuous on [a; b], there is � > 0 such that x , w 2 [a; b], jx � wj < � )j f .x/� f .w/j < "=.b� a/. Let P D fa D x0; x1; : : : ; xn D bg be a partition of [a; b] with max1� j�n

jx j � x j�1j < �. By

46

the extreme value theorem, Mj D f .wj / and m j D f .u j / for some wj; u j 2 [x j�1; x j]: Then

U . f; P/ � L. f; P/ D nXjD1

�f .wj/ � f .u j /�1x j < "

b � a

nXjD1

.x j � x j�1/ D ".xn � x0/b � a

D ":By the integral criterion, f is integrable on [a; b]:Exercises. (1) Let f : [a; b] ! Rbe a function and c 2 [a; b]: Show that f is integrable on [a; b] if and only if f isintegrable on [a; c] and [c; b]:(2) If f : [a; b] ! R is bounded and discontinuous only at x1; : : : ; xn 2 [a; b]; show that f is integrable on [a; b]:

(Hint: This can be done directly or by using (1) to reduce the problem to intervals having only one discontinuity.)

Questions. How bad can an integrable function be discontinuous? Which functions are integrable?

Definitions.(i) A set S � Ris of measure 0 (or has zero-length) iff for every " > 0, there are intervals .a1; b1/; .a2; b2/;.a3; b3/; : : : such that S � 1[nD1

.an; bn/ and1X

nD1

jan � bnj < ".

(ii) We say a property holds almost everywhere iff the property holds except on a set of measure 0. (It is common toabbreviate almost everywhere by a.e. in advanced courses. In probability, almost surely is used instead of almosteverywhere.)

Remarks. For example, it is known that monotone functions on intervals are differentiable almost everywhere (seeH.L. Royden’s book, Real Analysis, 3rd ed., p. 100). Also, H. Rademacher proved that Lipschitz functions on intervalsare differentiable almost everwhere (see L.C. Evans’ book, Partial Differential Equations, p. 281).

Lebesgue’s Theorem(1902). For a bounded function f : [a; b] ! R; f is integrable on [a; b] if and only if the setSf D fx 2 [a; b]: f is discontinuous at xg is of measure 0 (i.e. f is continuous almost everywhere).

For a proof of Lebesgue’s theorem, please go to appendix 1 of this chapter.

Examples. (1) The empty set ; is of measure 0 because ; � 1[nD1

.0; 0/ and1X

nD1

j0� 0j D 0 < ": So every continuous

function f : [a; b] ! R is integrable on [a; b]; since the set of discontinuities S is ;.

(2) A countable set fx1; x2; : : :g is of measure 0 becausefx1; x2; x3; : : :g � .x1 � "4; x1 C "

4/ [ .x2 � "

42; x2 C "

42/ [ .x3 � "

43; x3 C "

43/ [ � � � and

1XnD1

2"4n

D 2"3

< ":So every monotone function on [a; b] is integrable on [a; b]; since the set of discontinuities S is countable by themonotone function theorem, hence of measure 0.

(3) There exist uncountable sets, which are of measure 0 (eg. the Cantor set).

(4) A countable union of sets of measure 0 is of measure 0. (If S1; S2; S3; : : : are of measure 0 and S is their union,

then for every " > 0; since Sk is of measure 0, there are intervals .ak;n; bk;n/ such that Sk � 1[nD1

.ak;n; bk;n/ and1XnD1

jak;n � bk;nj < "4k: Then S � 1[

kD1

1[nD1

.ak;n ; bk;n/ and1X

kD1

1XnD1

jak;n � bk;nj < 1XkD1

"4kD "

3< ":)

(5) If a set S is of measure 0 and S0 � S; then S0 is also of measure 0. (For S0; use the same intervals .an; bn/ for S:)(6) Arrange Q\ [0; 1] in a sequence r1; r2; r3; : : : and define fn.x/ D n

1 if x D r1 or r2 or : : : or rn

0 otherwise: Then fn.x/

is integrable on [0; 1]; since the set of discontinuities Sfn D fr1; r2; : : : ; rng is finite, hence of measure 0. Now

47

limn!1 fn .x/ D f .x/ D �

1 if x 2 Q0 if x 62 Q is not integrable on [0; 1] since Sf D fx 2 [0; 1] : f is discontinuous at xg D

[0; 1];which is not of measure 0. So the limit of a sequence of Riemann integrable functions may not be Riemannintegrable.

Remarks. In B. R. Gelbaum and J. M. H. Olmsted’s book Counterexamples in Analysis, p. 106, there is anexample of the limit of a sequence of Acontinuous functions on [0; 1]; which is not Riemann integrable.

Theorem. For c 2 [a; b]; f is integrable on [a; b] if and only if f is integrable on [a; c] and [c; b]:Proof. Let S; S1; S2 be the set of discontinuous points of f on [a; b]; [a; c]; [c; b]; respectively. If f is integrableon [a; b]; then by Lebesgue’s theorem, S is of measure 0. Since S1; S2 � S; so both S1; S2 are of measure 0. ByLebesgue’s theorem, f is integrable on [a; c] and [c; b]:

Conversely, if f is integrable on [a; c] and [c; b]; then S1; S2 are of measure 0. Since S � S1 [ S2 [ fcg; byexamples (4) and (5), S is of measure 0. By Lebesgue’s theorem, f is integrable on [a; b]:Theorem. If f; g : [a; b] ! Rare integrable on [a; b]; then f C g; f � g and f g are integrable on [a; b]:Proof. If f and g are integrable on [a; b]; then f and g are bounded on [a; b]: So f C g; f � g and f g are boundedon [a; b]:

Observe that if f and g are continuous at x; then f C g is continuous at x : Taking contrapositive, we see that iff C g is discontinuous at x; then f or g is discontinuous at x : So if x 2 Sf Cg; then x 2 S f or x 2 Sg; which impliesSf Cg � S f [ Sg: Since f; g are integrable on [a; b]; by Lebesgue’s theorem, S f ; Sg are of measure 0. By example(4), Sf [ Sg is of measure 0. By example (5), SfCg is also of measure 0. Therefore, by Lebesgue’s theorem f C g isintegrable on [a; b]: Using a similar argument, we can see that f � g; f g are integrable on [a; b]:Remarks. By a similar argument, we can show that if f : [a; b] ! R is integrable on [a; b] and g is bounded andcontinuous on f .[a; b]/; then Sg� f � Sf : (This is because if f is continuous at x; then g � f is continuous at x : Takingcontrapositive, we get if x 2 Sg� f ; then x 2 Sf : So Sg� f � S f :) Since S f is of measure 0, by example (5), Sg� f isof measure 0. So g � f : [a; b] ! R is integrable on [a; b]: In particular, if f is integrable on [a; b]; then takingg.x/ D jxj; x2; ex ; cos x; : : : ; respectively, we get j f j; f 2; e f ; cos f; : : : are integrable on [a; b]:

However, even if f : [a; b] ! [c; d] is integrable on [a; b] and g : [c; d] ! R is integrable on [c; d]; g � f maynot be integrable on [a; b]: (For example, define f : [0; 1] ! [0; 1] by f .0/ D 1; f .m=n/ D 1=n; where m; n arepositive integers with no common prime factors, and f .x/ D 0 for every x 2 [0; 1]nQ: Next, define g : [0; 1] ! [0; 1]by g.0/ D 0 and g.x/ D 1 for every x 2 .0; 1]: As an exercise, it can be showed that Sf D [0; 1]\Qand Sg D f0g: Sof and g are integrable on [0; 1]:However, g � f is the nonintegrable function that is 1 on [0; 1]\Qand 0 on [0; 1]nQ:)

There is even an example of a continuous function f : [0; 1] ! [0; 1] and an integrable function g : [0; 1] ! Rsuch that g � f is not integrable (see B. R. Gelbaum and J. M. H. Olmstad’s book Counterexample in Analysis,pp. 106-107).

Up to now, we have been trying to determine which functions are integrable. Below we will look at how theintegrals of functions can be computed.

Theorem (Simple Properties of Riemann Integrals). Let f and g be integrable on [a; b]:(1)

Z b

a

�f .x/ � g.x/� dx DZ b

af .x/dx �Z b

ag.x/dx; Z b

acf .x/ dx D c

Z b

af .x/ dx for every c 2 R.

(2) If f .x/ � g.x/ for all x 2 [a; b], thenZ b

af .x/ dx � Z b

ag.x/ dx . Also,

����Z b

af .x/ dx

���� � Z b

aj f .x/j dx .

(3)Z b

af .x/ dx D Z c

af .x/ dx C Z b

cf .x/ dx for c 2 [a; b]:

Proof. To get (1), by the supremum and infimum properties, for every " > 0; there is a partition P of [a; b] such thatZ b

af .x/ dx C Z b

ag.x/ dx � " < L. f; P/ C L.g; P/ � L. f C g; P/ � Z b

a

�f .x/ C g.x/� dx

48

� U . f C g; P/ � U . f; P/ CU .g; P/ < Z b

af .x/ dx C Z b

ag.x/ dx C ":

Letting " ! 0; we getZ b

a

�f .x/ C g.x/� dx D Z b

af .x/ dx C Z b

ag.x/ dx : Next, from U .� f; P/ D �L. f; P/ and

inf.�S/ D � sup S; we getZ b

a� f .x/ dx D � Z b

af .x/ dx : ThenZ b

a

�f .x/ � g.x/� dx D Z b

a

�f .x/ C .�g.x//�dx D Z b

af .x/ dx C Z b

a�g.x/ dx D Z b

af .x/ dx � Z b

ag.x/ dx :

For the second statement, the case c D 0 is clear sinceZ b

a0 f .x/ dx D 0 D 0

Z b

af .x/ dx : If c > 0; thenZ b

ac f .x/dx D supfcL. f; P/: P is a partition of [a; b]gD c supfL. f; P/: P is a partition of [a; b]g D c

Z b

af .x/dx :

If c < 0; thenZ b

ac f .x/ dx D Z b

a�.�c/ f .x/ dx D � Z b

a.�c/ f .x/ dx D �.�c/ Z b

af .x/ dx D c

Z b

af .x/ dx :

For (2), observe that g � f � 0 on [a; b] implies L.g � f; P/ � 0 for every partition P: SoZ b

a

�g.x/ � f .x/� dx D supfL.g � f; P/: P is a partition of [a; b]g � 0 implies

Z b

af .x/ dx � Z b

ag.x/ dx :

For the second statement, since �j f j � f � j f j on [a; b]; we get � Z b

aj f .x/j dx � Z b

af .x/ dx � Z b

aj f .x/j dx :

For (3), let P be a partition of [a; b]; P 0 D P [ fcg; P1 D P 0 \ [a; c] and P2 D P 0 \ [c; b]: Then P 0 is a finerpartition of [a; b] than P; P1 is a partition of [a; c]; P2 is a partition of [c; b] and L. f; P 0/ D L. f; P1/C L. f; P2/: Let

A D fL. f; P/: P is a partition of [a; b]g and B D fL. f; P 0/: P is a partition of [a; b] and P 0 D P [ fcgg:Since P 0 is also a partition of [a; b]; we see B � A and so sup B � sup A: By the refinement theorem, L. f; P/ �L. f; P 0/ � sup B: Hence sup A � sup B: Therefore, sup A D sup B andZ b

af .x/ dx D sup A D sup BD supfL. f; P1/ C L. f; P2/: P1; P2 are partitions of [a; c]; [c; b]; respectivelygD supfL. f; P1/: P1 is a partition of [a; c]g C supfL. f; P2/: P2 is a partition of [c; b]gD Z c

af .x/ dx C Z b

cf .x/ dx :

The most important tool for computing an integral is to find an antiderivative or primitive function of the integrablefunction, which is a function whose derivative is the integrable function. What can we say about such a function?

Example. For x 2 [�1; 1]; define f .x/ D n 1 if x � 0�1 if x < 0: Since f .x/ has only one point of discontinuity at 0, f .x/

is integrable on [�1; 1]: Now the function

F.x/ D Z x

0f .t/ dt D n

x if x � 0�x if x < 0D jxj

49

is continuous on [�1; 1]; hence uniformly continuous there by the uniform continuity theorem, but F.x/ is notdifferentiable at 0, the same point where f .x/ is discontinuous!

Theorem. If f is integrable on [a; b] and c 2 [a; b]; then F.x/ D Z x

cf .t/ dt is uniformly continuous on [a; b]:

Proof. Recall there is K > 0 such that j f .t/j � K for every t 2 [a; b]: For every " > 0; let � D "K : Then jx � x0j < �

implies jF.x/ � F.x0/j D ����Z x

x0

f .t/ dt

���� � K jx � x0j < ":The next theorem is the most important theorem in calculus. It not only tells us how to compute an integral,

but also the deep fact that differentiation and integration are inverse operations on functions. Roughly, it may be

summarized by the formulasd

dx

Z x

ch.t/ dt D h.x/ and

Z x

c

d

dth.t/ dt D h.t/���x

c:

Fundamental Theorem of Calculus. Let c; x0 2 [a; b]:(1) If f is integrable on [a; b], continuous at x0 2 [a; b] and F.x/ D Z x

cf .t/ dt , then F 0.x0/ D f .x0/.

(2) If G is differentiable on [a; b] with G0 D g integrable on [a; b], thenZ b

ag.x/ dx D G.b/ � G.a/.

(Note G0 need not be continuous by an example in the chapter on differentiation.)

Proof. (1) We will check limx!x0

F.x/ � F.x0/x � x0

D f .x0/ using the definition of limit. For every " > 0; since f is

continuous at x0; there exists a � > 0 such that for every x 2 [a; b]; jx � x0j < � ) j f .x/ � f .x0/j < "=2: With thesame �; we get 0 < jx � x0j < � implies���� F.x/ � F.x0/

x � x0� f .x0/���� D ����� R x

c f .t/dt � R x0

c f .t/dt

x � x0� R x

x0f .x0/dt

x � x0

����� D ����� R xx0. f .t/� f .x0//dt

x � x0

����� � ����� R xx0"=2 dt

x � x0

����� < ":(2) The conclusion will follow from the infinitesimal principle if we can show

���Z b

ag.x/ dx � �G.b/ � G.a/���� < "

for every " > 0: Now for every " > 0; since g is integrable on [a; b]; by the integral criterion, there is a partitionP D fa D x0; x1; : : : ; xn D bg such that U .g; P/ � L.g; P/ < ": For j D 1; 2; : : : ; n; by the mean value theorem,there exists tj 2 .x j�1; x j/ such that G.x j / � G.x j�1/ D G0.tj /.x j � x j�1/ D g.tj/1x j : Then

L.g; P/ � nXjD1

g.ti/1x j D nXjD1

.G.x j / � G.x j�1// D G.b/ � G.a/ � U .g; P/:Also, L.g; P/ � Z b

ag.x/ dx � U .g; P/: Then

���Z b

ag.x/ dx � �G.b/ � G.a/���� � U .g; P/ � L.g; P/ < ":

Remarks. In (2) above, the condition G0 is integrable is important. The function G.x/ D x2 sin1x2

for x 6D 0 and

G.0/ D 0 is differentiable on R; but G0.x/ is unbounded on [0; 1]; hence not integrable there. In Y. Katznelson andK. Stromberg’s paper Everywhere Differentiable, Nowhere Monotone Functions, American Mathematical Monthly,vol. 81, pp. 349-354, there is even an example of a differentiable function onRsuch that its derivative is bounded, butnot Riemann integrable on any interval [a; b] with a < b:Theorem (Integration by Parts). If f , g are differentiable on [a; b] with f 0, g0 integrable on [a; b], thenZ b

af .x/g0.x/ dx D f .b/g.b/� f .a/g.a/ � Z b

af 0.x/g.x/ dx :

50

Proof. By part (2) of the fundamental theorem of calculus,Z b

a. f g/0.x/ dx D f .b/g.b/� f .a/g.a/: Since . f g/0.x/ D

f 0.x/g.x/C f .x/g0.x/; we can subtract the integral of f 0.x/g.x/ on both sides to get the integration by parts formula.

Theorem (Change of Variable Formula). If � : [a; b] ! R is differentiable, �0 is integrable on [a; b] and f is

continuous on ��[a; b]�, then

Z �.b/�.a/ f .t/ dt D Z b

af��.x/��0.x/ dx :

Proof. Let g.x/ D Z �.x/�.a/ f .t/ dt : By part (1) of the fundamental theorem of calculus and chain rule, g0.x/ Df��.x/��0.x/: So

Z b

af��.x/��0.x/ dx D Z b

ag0.x/ dx D g.b/� g.a/ D g.b/ D Z �.b/�.a/ f .t/ dt :

Appendix 1: Proof of Lebesgue’s Theorem

We first modify the uniform continuity theorem to obtain a lemma.

Lemma. Let f : [a; b] ! R be a function continuous on a subset K D [a; b] n 1[jD1

.�j ; �j/: Then for every " > 0;there exists a � > 0 such that for all x 2 K ; t 2 [a; b]; jx � t j < � implies j f .x/ � f .t/j < ":Proof. Suppose the lemma is false. Then there is an " > 0 such that for every � D 1

n ; n 2 N; there are xn 2 K ; tn 2 [a; b]such that jxn � tnj < � D 1

n and j f .xn/ � f .tn/j � ": By the Bolzano-Weierstrass theorem, fxng has a subsequencefxnj g converging to some w 2 [a; b]: Sincejtnj � wj � jtnj � xnj j C jxnj � wj � 1n jC jxnj � wj;

by the sandwich theorem, ftnj g converges to w:Next we will show w 2 K : Suppose w 62 K ; then w 2 .�i ; �i/ for some i: Since lim

j!1 xnj D w and w 2 .�i ; �i /;by the definition of limit, there will be some xnp 2 .�i ; �i /; contradicting xnp 2 K D [a; b] n 1[

jD1.�j ; �j/: So w 2 K :

Since f is continuous at w; by the sequential continuity theorem,

0 D j f .w/ � f .w/j D limj!1 j f .xnj / � f .tnj /j � " > 0;

which is a contradiction.

Proof of Lebesgue’s Theorem. First suppose f : [a; b] ! R is integrable. Note that if f is discontinuous at x; thenthere is an " > 0 such that for every � > 0; there is z 2 .x��; xC�/\[a; b] such that j f .x/� f .z/j � ": Let Dk be the

set of all x 2 [a; b] such that for every open interval I with x 2 I; there is z 2 I \ [a; b] such that j f .x/ � f .z/j > 1

k:

Since every positive " > 1

kfor some positive integer k by the Archimedian principle and every open interval I with x 2 I

contains an interval .x��; xC�/ for some � > 0; it follows that Sf D fx 2 [a; b] : f is discontinuous at xg D 1[kD1

Dk :We will show each Dk is of measure 0, which will imply Sf is of measure 0.

For every " > 0; by the integral criterion, there is a partition P D fx0; x1; : : : ; xng of [a; b] such that

U . f; P/ � L. f; P/ < "2k

: If there is x 2 Dk \ .x j�1; x j/; then Mj � m j � j f .x/ � f .z/j > 1

kfor some z 2.x j�1; x j/: Let J be the set of j such that Dk \ .x j�1; x j/ 6D ;: Then Dk n fx0; x1; : : : ; xng � [

j2J.x j�1; x j/ andX

j2J

jx j�1 � x j j �Xj2J

k .Mj �m j /| {z }>1=k

1x j � k�U . f; P/ � L. f; P/� < "

2: Next around each x j;we take an open interval

51

Ij containing x j with length"

2.n C 1/ : Then the intervals .x j�1; x j/ with j 2 J and Ij ’s together contain Dk and the

sum of their lengths is less than ": So each Dk (and hence Sf ) is of measure 0.

Conversely, suppose Sf is of measure 0. For every " > 0; let "0 D "3.N C b � a/ ; where N is an upper

bound for j f .x/j on [a; b]: By the definition of measure 0, there are intervals .�i ; �i / such that Sf � 1[iD1

.�i ; �i /and

1XiD1

j�i � �i j < "0: Then f is continuous on K D [a; b] n 1[iD1

.�i ; �i /: Now take the � in the lemma for "0 and

a partition P D fx0; x1; : : : ; xng such that jx j�1 � x j j < � for j D 1; 2; : : : ; n: If K \ [x j�1; x j] D ;; then

[x j�1; x j] � 1[iD1

.�i ; �i /: If there is x 2 K \ [x j�1; x j]; then Mj � m j D supfj f .t0/ � f .t1/j : t0; t1 2 [x j�1; x j]g �supfj f .t0/ � f .x/j C j f .x/ � f .t1/j : t0; t1 2 [x j�1; x j]g � 2"0: So

U . f; P/ � L. f; P/ D XK\[xj�1;xj ]D; .Mj � m j /| {z }�2N

1x j C XK\[xj�1;xj ] 6D; .Mj � m j /| {z }�2"0

1x j � 2N"0 C 2"0.b � a/ < ":Therefore, by the integral criterion, f is integrable on [a; b]:

Appendix 2: Riemann’s Definition of the Integral

Recall the definition of limx!0

f .x/ D L is that for every " > 0; there is � > 0 such that for every x with jxj < �;we have j f .x/ � L j < ": Based on this, Riemann’s approach of integral can be defined as follow.

Definition. Let f be a bounded function on [a; b]: We write limkPk!0

nXjD1

f .tj/1x j D I iff for every " > 0; there is a� > 0 such that for every partition P of [a; b] with kPk < � and every choice tj 2 [x j�1; x j]; j D 1; 2; : : : ; n; we

have��� nX

jD1

f .tj/1x j � I��� < ": (Such I is unique as in the proof of uniqueness of limit.)

Below we will show Darboux’s approach is the same as Riemann’s by establishing the following theorem.

Darboux’s Theorem. Let f be a bounded function on [a; b]: Then the following are equivalent:

(a)Z b

af .x/ dx D I I

(b) limkPk!0

nXjD1

f .tj /1x j D I I(c) for every " > 0; there is a partition P of [a; b] such that for every refinement Q D fx0; x1; : : : ; xng of P; every

choice tj 2 [x j�1; x j]; we have��� nX

jD1

f .tj/1x j � I��� < ": (Such a number I is unique as in the proof of uniqueness of

limit.)

Proof. .a/ ) .b/ SupposeZ b

af .x/ dx D I: For every " > 0; by the proof of the integral criterion, there is a

partition P" D fw0; w1; : : : ; wqg of [a; b] such that I � "2< L. f; P"/ � U . f; P"/ < I C "

2: Let � D "

4q K; where

K D supfj f .x/j : x 2 [a; b]g: If P D fx0; x1; : : : ; xng is a partition of [a; b] with kPk < �; then U . f; P/ D S1 C S2;where S1 is the sum of the terms from intervals not containing points in P" and S2 the sum of the remaining terms.

Note the rectangles for the terms of S1 are inside the rectangles for U . f; P"/: So we have S1 < U . f; P"/ < I C "2:

Since P" D fw0; w1; : : : ; wqg; S2 has at most 2q terms (one for w0; wq each and at most two for w1; : : : ; wq�1

52

each). Then S2 � 2q.KkPk/ � 2q K� D "2: Thus U . f; P/ D S1 C S2 < I C ": Similarly, L. f; P/ > I � ": So

I � " < L. f; P/ � nXjD1

f .tj/1x j � U . f; P/ < I C ": Then��� nX

jD1

f .tj /1x j � I��� < ":.b/ ) .c/ For every " > 0; take � as the definition above. Let P be any partition of [a; b] with kPk < �: Then for

every refinement Q of P; we have kQk � kPk < � and so (c) follows from the conclusion of (b)..c/ ) .a/ For " > 0; let P be the partition in (c) for"3: For each j D 1; 2; : : : ; n; by the supremum limit theorem, there

are sequences tj;k such that limk!1 f .tj;k/ D Mj D supf f .x/ : x 2 [x j�1; x j]g: Since

��� nXjD1

f .tj;k/1x j � I��� < "

3; so taking

limit, we get jU . f; P/ � I j � "3: Similarly, we can also get jL. f; P/ � I j � "

3: Then jU . f; P/ � L. f; P/j � 2"

3< ":

By the integral criterion, it follows thatZ b

af .x/ dx exists. By .a/) .b/ and .b/) .c/; it must equal I by uniqueness.

Improper Riemann Integral – Integration of Unbounded Functions or on Unbounded Intervals

We would like to integrate unbounded functions on unbounded intervals by taking limit of integrals on boundedintervals. Since the functions may or may not be continuous, we have to make sure the functions are integrable onbounded intervals first.

Definition. Let I be an interval. We say f : I ! R is locally integrable on I iff f is integrable on every closed andbounded subintervals of I: We denote this by f 2 L loc.I /.

Roughly, there are three cases of improper integrals.

Case 1: (Unbounded near one endpoint) Let f be locally integrable on [a; b/; where a is real and b is real or C1:We define the improper (Riemann) integral of f on [a; b/ to be

Z b

af .x/ dx D lim

d!b� Z d

af .x/ dx; provided the limit

exists in R; and we say f is improper integrable on [a; b/ in that case. For f locally integrable on an interval of theform .a; b]; where a is real or �1 and b is real, the definitions of the improper (Riemann) integral of f on .a; b] andf is improper (Riemann) integrable on .a; b] are similar.

Case 2: (Unbounded near both endpoints) Let f be locally integrable on .a; b/;with a; b real or infinity, and x0 2 .a; b/.We define the improper (Riemann) integral of f on .a; b/ to be

Z b

af .x/ dx D lim

c!aC Z x0

cf .x/ dx C lim

d!b� Z d

x0

f .x/ dx

provide both limits exist in R; and say f is improper integrable on .a; b/ in that case. (Note the integral is the sameregardless of the choices of x0 2 .a; b/.)Case 3: (Unbounded inside interval) Let f be improper integrable on intervals [a; c/ and .c; b]: The improper integralof f on [a; b] is the sum of the improper integrals on the two intervals. The cases where a or b is excluded are similarlydefined.

In each case, if the improper integral is a number, then we say the improper integral converges to the number,otherwise we say it diverges.

Examples.(1) Consider the unbounded function f .x/ D ln x on .0; 1]. Observe that f is continuous, hence integrable,

on every closed subinterval of .0; 1]: So f 2 L loc.0; 1]: Now limc!0C Z 1

cln x dx D lim

c!0C.�1� c ln cC c/ D �1, soZ 1

0ln x dx D �1 and ln x is improper integrable on .0; 1]:

53

(2) Consider f .x/ D 1

x2on the unbounded interval [2;C1/. Observe that f 2 L loc[2;C1/ because of continuity.

Now limd!C1Z d

2

1

x2dx D lim

d!C1�� 1

dC 1

2

� D 1

2, so

Z 12

1

x2dx D 1

2and

1

x2is improper integrable on [2;C1/:

(3) Consider f .x/ D ex onRD .�1;C1/: Since f is continuous everywhere, f 2 L loc.�1;C1/: Now take a

x0 2 .�1;C1/; say x0 D 0: Then limc!�1Z 0

cex dx D lim

c!�1.1 � ec/ D 1; but limd!C1Z d

0ex dx D lim

d!C1.ed � 1/does not exist inR: So ex is not improper integrable on .�1;C1/:

Remarks. By taking limits, many properties of Riemann integrals extend to improper Riemann integrals as well.There are also some helpful tests to determine if a function is improper integrable.

p-test. For 0 < a <1; we haveZ 1

a

1x p

dx <1 if and only if p > 1: Also,Z a

0

1x p

dx <1 if and only if p < 1:Proof. In the proof of p-test for series, we saw

Z 11

1x p

dx <1 if and only if p > 1: AddingZ 1

a

1x p

dx <1; we

get the first statement. Next, by the change of variable y D 1

x; we have

Z 1

c

1

x pdx D Z 1=c

1

1

y2�pdy: Taking limit as

c ! 0C; we see thatZ 1

0

1

x pdx <1 if and only if

Z 11

1

y2�pdy <1: By the first statement, this is the same as

2� p > 1; i.e. p < 1: Again addingZ a

1

1x p

dx <1; we get the second statement.

Theorem (Comparison Test). Suppose 0 � f .x/ � g.x/ on I and f; g 2 L loc.I /: If g is improper integrable on I ,then f is improper integrable on I . If f is not improper integrable on I , then g is not improper integrable on I .

Proof. For the case I D [a; b/; if 0 � f � g on I and g is improper integrable on I; thenZ d

af .x/ dx is an increasing

function of d and is bounded above byZ b

ag.x/ dx : So

Z b

af .x/ dx D lim

d!b� Z d

af .x/ dx exists inRby the monotone

function theorem. The other cases I D .a; b] and .a; b/ are similar.

Theorem (Limit Comparison Test). Suppose f .x/; g.x/ > 0 on .a; b] and f; g 2 L loc..a; b]/: If limx!aC g.x/

f .x/ is a

positive number L ; then either (bothZ b

af .x/ dx and

Z b

ag.x/ dx converge) or (both diverge). If lim

x!aC g.x/f .x/ D 0;

thenZ b

af .x/ dx converges implies

Z b

ag.x/ dx converges. If lim

x!aC g.x/f .x/ D 1; then

Z b

af .x/ dx diverges impliesZ b

ag.x/ dx diverges. In the case [a; b/; we take lim

x!b� g.x/f .x/ and the results are similar.

Proof. If limx!aC g.x/

f .x/ is a positive number L ; then for " D L

2> 0; there is � > 0 such that for every x 2 .a; aC �/; we

haveL

2D L � " < g.x/

f .x/ < L C " D 3L

2: Then

L

2

Z aC�a

f .x/ dx � Z aC�a

g.x/ dx � 3L

2

Z aC�a

f .x/ dx : So either

(bothZ b

af .x/ dx and

Z b

ag.x/ dx converge) or (both diverge).

If limx!aC g.x/

f .x/ D 0; then there exists �0 > 0 such that for every x 2 .a; a C �0/; we have 0 < g.x/f .x/ < 1; which

implies 0 < g.x/ < f .x/: Then 0 � Z aC�0a

g.x/ dx � Z aC�0a

f .x/ dx : SoZ b

af .x/ dx <1) Z b

ag.x/ dx <1:

54

If limx!aC g.x/

f .x/ D 1; then there is �00 > 0 such that for every x 2 .a; a C �00/; we haveg.x/f .x/ > 1; which implies

g.x/ > f .x/: HenceZ aC�00

ag.x/ dx � Z aC�00

af .x/ dx : So

Z b

af .x/ dx D 1) Z b

ag.x/ dx D 1:

Theorem (Absolute Convergence Test). If f 2 L loc.I / and j f j is improper integrable on I , then f is improperintegrable on I .

Proof. We have �j f j � f � j f j on I; which implies 0 � f C j f j � 2j f j on I: By the comparison test, f C j f j isimproper integrable on I: Therefore, f D . f C j f j/ � j f j is improper integrable on I:Examples. (4) Is

ln x

1C x2improper integrable on .0; 1]? Observe first that the function is locally integrable on .0; 1]

by continuity. Also,

���� ln x

1C x2

���� � j ln xj on .0; 1]: Since j ln xj D � ln x is improper integrable on .0; 1] (cf.

example (1)), so

���� ln x

1C x2

���� is improper integrable on .0; 1] by the comparison test. Thenln x

1C x2is improper

integrable on .0; 1] by the absolute convergence test.

(5) DoesZ C1

2

dxpx2 � 1

converge? Observe that on [2;C1/; 0 < 1

x� 1p

x2 � 1: Both

1

xand

1px2 � 1

are

continuous, hence locally integrable, on [2;C1/: NowZ C1

2

1

xdx D1 by p-test. By the comparison test,Z C1

2

dxpx2 � 1

diverges.

(6) DoesZ C1

1

sin x

xdx converge? Since

sin x

xis continuous on [1;C1/; it is locally integrable there. Integrating

by parts, Z c

1

sin x

xdx D � cos c

cC cos 1� Z c

1

cos x

x2dx :

Since j cos cj � 1; limc!C1 cos c

cD 0: Since

���cos x

x2

��� � 1

x2on [1;C1/ and

Z C11

1

x2dx <1 by p-test, we haveZ C1

1

cos x

x2dx converges by the comparison test and the absolute convergence test. Therefore,

Z C11

sin x

xdx

converges.

(7) For the improper integralZ 1

0

dx

1� x3; the integrand is continuous (hence locally integrable) on [0; 1/:Observe that

11� x3

D 11� x

� 11C x C x2

�and the second factor on the right has a positive limit as x ! 1� : More precisely,

since limx!1� 1=.1� x/

1=.1� x3/ D limx!1�.1 C x C x2/ D 3 and

Z 1

0

dx

1� xD lim

d!1�Z d

0

dx

1� xD lim

d!1�� ln.1� d/ D 1;by the limit comparison test,

Z 1

0

dx

1� x3diverges.

(8) For the improper integralZ 5

0

dx3p

7x C 2x4; the integrand is continuous(hence locally integrable) on .0; 5]:Observe

that1

3p

7x C 2x4D 1

3p

x

� 13p

7C 2x3

�and the second factor on the right has a positive limit as x ! 0C : More

precisely, since limx!0C 1= 3

px

1= 3p

7x C 2x4D 3p

7 andZ 5

0

dx3p

x<1 by p-test as p D 1

3< 1; by the limit comparison

test,Z 5

0

dx3p

7x C 2x4converges.

55

Principal Value – Symmetric Integration about Singularities.

Definition. Let f 2 L loc.R/; then the principal value ofZ 1�1 f .x/ dx is P:V: Z 1�1 f .x/ dx D lim

c!C1Z c�cf .x/ dx .

Examples. (1) Because of continuity, f .x/ D 11C x2

2 L loc.R/: Now P:V: Z 1�1 11C x2

dx D limc!C1Z c�c

11C x2

dxD limc!C1.2 tan�1 c/ D � exists and the improper integralZ 1�1 1

1C x2dx D lim

c!�1Z 0

c

11C x2

dx C limd!C1Z d

0

11C x2

dx D limc!�1.� tan�1 c/ C lim

d!C1.tan�1 d/ D �:(2) Because of continuity, f .x/ D x 2 L loc.R/: Now P:V: Z 1�1 x dx D lim

c!C1Z c�cx dx D lim

c!C10 D 0 exists, how-

ever the improper integralZ 1�1 x dx D lim

c!�1Z 0

cx dx C lim

d!C1Z d

0x dx D lim

c!�1.�c2

2/ C lim

d!C1.d2

2/ does not

exist.

Theorem. If the improper integralZ 1�1 f .x/ dx exists, then P:V: Z 1�1 f .x/ dx exists and equals the improper integralZ 1�1 f .x/ dx. (Note the converse is false by example (2).)

Proof. If the improper integralZ 1�1 f .x/ dx exists, then it implies that both lim

d!�1Z 0

df .x/ dx and lim

c!C1Z c

0f .x/ dx

exist. So

P:V: Z 1�1 f .x/ dx D limc!C1Z c�c

f .x/ dx D limc!C1�Z 0�c

f .x/ dx C Z c

0f .x/ dx

�D limd!�1Z 0

df .x/ dx C lim

c!C1Z 10

f .x/ dx D Z 1�1 f .x/ dx :Definition. For a; b 2 Ror 1; let f : [a; c/[ .c; b] ! Rbe locally integrable on [a; c/ and on .c; b]: Define

P:V: Z b

af .x/ dx D lim"!0C �Z c�"

af .x/ dx C Z b

cC" f .x/ dx

�:Example. Consider the improper sense and the principal value sense of

Z 1�1

1x

dx :Because of continuity, 1

x 2 L loc[�1; 0/ and 1x 2 L loc.0; 1]: In the improper sense,

Z 1

0

1

xdx D lim

c!0C Z 1

c

1

xdxD lim

c!0C.� ln c/ does not exist. So the improper integral on [�1; 0/ [ .0; 1] does not exist. However,

P:V: Z 1�1

1x

dx D lim"!0C�Z �"�1

1x

dx C Z 1" 1x

dx

� D lim"!0C.ln j � "j � ln j"j/ D 0:Remarks. (1) It is incorrect to say

Z 1�1

1x

dx D ln j � 1j � ln j1j D 0;which attempts to use part (2) of the fundamental

theorem of calculus, but here f .x/ D ln jxj is not differentiable on [�1; 1] failing the required condition.

(2) There is a similar theorem for this type of principal value integrals as the theorem above.

56

Taylor Series of Common Functions

Question. How do calculators compute sin x; cos x; ex; x y; ln x; : : :?Recall Taylor’s theorem with the Lagrange form of the remainder is the following.

Taylor’s Theorem. Let f : .a; b/! Rbe n times differentiable on .a; b/. For every x, c 2 .a; b/, there is x0 betweenx and c such that

f .x/ D f .c/ C f 0.c/1!

.x � c/ C f 00.c/2!

.x � c/2 C : : : C f .n�1/.c/.n � 1/! .x � c/n�1 C f .n/.x0/n!

.x � c/n :(This is called the n-th Taylor expansion of f about c: The term Rn.x/ D f .n/.x0/

n!.x � c/n is called the Lagrange

form of the remainder.)

There are two other forms of the remainder.

Theorem (Taylor’s Formula with Integral Remainder). Let f be n times differentiable on .a; b/: For everyx; c 2 .a; b/; if f .n/ is integrable on the closed interval with endpoints x and c; then

f .x/ D f .c/C f 0.c/1!

.x�c/C� � �C f .n�1/.c/.n � 1/! .x�c/n�1CRn .x/; where Rn.x/ D 1.n � 1/! Z x

c.x�t/n�1 f .n/.t/ dt :

Proof. Noted

dt

��.x � t/� D 1: Applying integration by parts n � 1 times, we get

f .x/ � f .c/ D Z x

cf 0.t/ � 1 dt D Z x

cf 0.t/��.x � t/�0 dtD � f 0.t/.x � t/���x

cC Z x

c.x � t/ f 00.t/ dtD f 0.c/.x � c/ C �� f 00.t/ .x � t/2

2!

���xcC 1

2!

Z x

c.x � t/2 f 000.t/ dt

� D � � �D f 0.c/.x � c/ C � � � C f .n�1/.c/.n � 1/! .x � c/n�1 C 1.n � 1/! Z x

c.x � t/n�1 f .n/.t/ dt :

Mean Value Theorem for Integrals. If f is continuous on [a; b], thenZ b

af .x/ dx D f .x1/.b � a/ for some

x1 2 [a; b], (i.e. the mean value or average of f on [a; b] is1

b� a

Z b

af .x/ dx D f .x0/). More generally, if g is

integrable and g � 0 on [a; b]; thenZ b

af .x/g.x/ dx D f .x1/ Z b

ag.x/ dx for some x1 2 [a; b].

Proof. The first statement is the special case of the second statement when g.x/ D 1: So we only need to prove thesecond statement. Let M and m be the maximum and the minimum of f on [a; b]; respectively. Since m � f .x/ � M

on [a; b]; we have mZ b

ag.x/ dx � Z b

af .x/g.x/ dx � M

Z b

ag.x/ dx : If

Z b

ag.x/ dx D 0; then the last sentence

impliesZ b

af .x/g.x/ dx D 0 and so we may take x1 to be any element of [a; b]: If

Z b

ag.x/ dx > 0; then dividing byZ b

ag.x/ dx; we see that

Z b

af .x/g.x/ dx

,Z b

ag.x/ dx is between m and M: By the intermediate value theorem,

this expression equals f .x1/ for some x1 2 [a; b]; which gives the second statement.

57

Theorem (Taylor’s Formula with Cauchy Form Remainder). Let f be n times differentiable on .a; b/: Forx; c 2 .a; b/; if f .n/ is continuous on the closed interval with endpoints x and c; then there is x1 between x and c suchthat

f .x/ D f .c/ C f 0.c/1!

.x � c/C � � � C f .n�1/.c/.n � 1/! .x � 1/n�1 C Rn.x/; where Rn.x/ D .x � c/.x � x1/n�1 f .n/.x1/.n � 1/! :Proof. This follows by applying the mean value theorem for integrals to the integral remainder of Taylor’s formulaabove.

Remarks. For every x 2 R; the series1X

kD0

xk

k!converges. This follows from the ratio test because lim

k!1 ���� xkC1.k C 1/! k!

xk

����D limk!1 jxj

k C 1D 0 < 1: By term test, we have lim

k!1 xk

k!D 0 for every x 2 R:

Examples.

(1) For f .x/ D sin x , f .n/.x/ D � .�1/k cos x if n D 2k C 1.�1/k sin x if n D 2k. So j f .n/.x/j � 1 for every x 2 R:

Taking c D 0; we have f .2kC1/.0/ D .�1/k and f .2k/.0/ D 0 for every k 2 N: By Taylor’s theorem,

sin x D n�1XkD0

.�1/k x2kC1.2k C 1/! C R2n.x/ for every x 2 R. Now jR2n.x/j � 1.2n/! jxj2n ! 0 as n ! 1 by the re-

marks above. Therefore,

sin x D x � x3

3!C x5

5!� x7

7!C � � � D 1X

kD0

.�1/k x2kC1.2k C 1/! for �1 < x < C1:Remarks. For 0 � x � �

2; jR18.x/j � jxj18

18!� .�=2/18

18!< 6� 10�13: So x � x3

3!C � � � C x17

17!can be used to

compute sin x to 10 decimal places.

(2) For f .x/ D cos x , f .n/.x/ D � .�1/k sin x if n D 2k � 1.�1/k cos x if n D 2k. So j f .n/.x/j � 1 for every x 2 R:

Taking c D 0; we have f .2k�1/.0/ D 0 and f .2k/.0/ D .�1/k for every k 2 N: By Taylor’s theorem,

sin x D n�1XkD0

.�1/k x2k.2k/! C R2n�1.x/ for every x 2 R. Now jR2n�1.x/j � 1.2n � 1/! jxj2n�1 ! 0 as n ! 1by the remarks above. Therefore,

cos x D 1� x2

2!C x4

4!� x6

6!C � � � D 1X

kD0

.�1/k x2k.2k/! for �1 < x < C1:(3) For f .x/ D ex , f .n/.x/ D ex : Taking c D 0; we have f .n/.0/ D 1 for every n 2 N: By Taylor’s theorem,

ex D n�1XkD0

xk

k!C Rn.x/ for every x 2 R: Now jRn.x/j D ex0

n!jxjn � max.e0; ex /

n!jxjn ! 0 as n ! 1 by the

remarks above. So

ex D 1C x C x2

2!C x3

3!C � � � D 1X

kD0

xk

k!for �1 < x < C1:

Remarks. The Taylor series for sin x; cos x and ex also converge if x is a complex number! In fact, this is howthe sine, cosine, and exponential functions are defined for complex numbers. For every x 2 R; we have

eix D 1XkD0

.ix/k

k!D 1X

nD0

.�1/n x2n.2n/! C i1X

nD0

.�1/n x2nC1.2n C 1/! D cos x C i sin x :58

In particular, we have ei� D cos� C i sin� D �1 so that ei� C 1 D 0: This is known as Euler’s formula. It isthe most beautiful formula in mathematics because it connects the five most important constants 1; 0; �; i; e ofmathematics in one single equation!

(4) (Binomial Theorem) For a 2 R; if �1 < x < 1; then.1 C x/a D 1C ax C a.a � 1/2!

x2 C a.a � 1/.a � 2/3!

x3 C � � � D 1C 1XkD1

a.a � 1/ � � � .a � k C 1/k!

xk :To see this, let f .x/ D .1 C x/a ; then f .n/.x/ D a.a � 1/ � � � .a � n C 1/.1 C x/a�n : By Taylor’s formula withintegral formula,

Rn.x/ D 1.n � 1/! Z x

0.x � t/n�1 f .n/.t/ dt D a.a � 1/ � � � .a � n C 1/.n � 1/! Z x

0

� x � t

1C t

�n�1.1C t/a�1 dt :For x 2 [0; 1/; the function g.t/ D x � t

1C thas derivative g0.t/ D �1 � x.1C t/2

< 0: On [0; x]; g.t/ � g.0/ D x : Let

k D Z x

0.1C t/a�1 dt; then jRn.x/j � ja.a � 1/ � � � .a � n C 1/kxn�1 j.n � 1/!| {z }

call this bn

: Since limn!1 bnC1

bnD lim

n!1 ja � njjxjn

Djxj < 1; by ratio test,1X

nD1

bn converges. By term test, limn!1 bn D 0: Then lim

n!1 Rn.x/ D 0 by the sandwich

theorem. So the binomial formula is true for x 2 [0; 1/: The case x 2 .�1; 0] is similar.

Here are a few more common Taylor series. (Note the series only equal the functions on a small interval.) They

are obtained from the cases a D �1; a D �1 (with x replaced by x2) and a D �1

2(with x replaced by �x2) of the

binomial theorem by term-by-term integration.

ln.1C x/ D x � x2

2C x3

3� x4

4C � � � D 1X

kD1

.�1/k�1xk

kfor � 1 < x � 1

tan�1 x D x � x3

3C x5

5� x7

7C � � � D 1X

kD0

.�1/k x2kC1

2k C 1for � 1 � x � 1

sin�1 x D x C 1

2

x3

3C 1 � 3

2 � 4

x5

5C � � � D x C 1X

kD1

1 � 3 � 5 � � � .2k � 1/2 � 4 � 6 � � � .2k/ x2kC1

2k C 1for � 1 � x � 1

Remarks. Unfortunately, the Taylor series of a function does not always equal to the function away from the center.

For example, the function f .x/ D �e�1=x2

if x 6D 00 if x D 0

can be shown to be infinitely differentiable and f .n/.0/ D 0

for every n 2 N: So the Taylor series of f .x/ about c D 0 is1X

kD0

0xk D 0; i.e. the Taylor series is the zero function.

Therefore, the Taylor series of f .x/ equals f .x/ only at the center c D 0:59

Practice Exercises

The starred problems are difficult and involve more work or deeper thinking.

For exercises 1 to 7, negate each of the following expressions or statements.

1. (x > 0 and x < 1) or x D �1

2. x > 0 and (x < 1 or x D �1)

3. For every triangle ABC; \A C\B C \C D 180�:4. There exists a man who does not have any wife.

5. For each x , there is a y such that x C y D 0:6. 9� 8� 9 such that j� � �j < :7. If x and y are positive, then x C y > 0.

8. Give the contrapositive of the following statements. (Since the contrapositives are equivalent to the statements,they say the same thing.)

(a) If AB D AC in 4ABC; then \B D \C in 4ABC:(b) If a function is differentiable, then it is continuous.

(c) If limx!0

f .x/ D a and limx!0

g.x/ D b; then limx!0

. f .x/ C g.x// D a C b:(d) If x2 C bx C c D 0; then x D �b Cp

b2 � 4c

2or x D �b �p

b2 � 4c

2:

9. Compute the following sets.

(a)�fx; y; zg [ fw; zg� n fu; v; wg: (Here u; v; w; x; y; z are distinct objects.)

(b) f1; 2g � f3; 4g � f5g:(c)Z\ [0; 10]\ fn2 C 1 : n 2 Ng:(d) fn 2 N : 5 < n < 9g n f2m : m 2 Ng:(e) .[0; 2] n [1; 3]/ [ .[1; 3] n [0; 2]/:

10. (i) Let A D [0; 1] and B D [0; 1][ [2; 3]: Plot the graphs of A � A and B � B on the plane.

(ii) If A; B are sets that are not the empty set and A� B D B � A; what can be said about A and B?

11. (a) If B � C; then prove that A [ B � A [ C:(b) For sets X; Y; Z ; prove that .X n Y / n Z D .X n Z / n Y:

12. (i) For all sets A; B;C; is it always true that .A [ B/ \ C D A [ .B \ C/?(ii) For all sets A; B;C; is it always true that if A [ B D A [ C; then B D C?

(iii) For all sets A; B;C; is it always true that A n .B [ C/ D .A n B/ \ .A n C/?13. (i) Show that if A � B and C � D; then A [ C � B [ D:

60

(ii) Is it always true that if A � B and C � D; then A [ C � B [ D?

(iii) For a < b; let.a; b/Q D fx : x 2 Q and a < x < bg and [a; b/Q D fx : x 2 Q and a � x < bg:Does

1[nD1

[1

n; 2/Q D 1[

nD1

.1

n; 2/Q?

14. Define functions f; g : R! Rby f .x/ D �0 if x > 01 if x � 0

and g.x/ D 1� 2x : For f and g; determine if each is

injective or surjective. Compute f � g and g � f:15. (i) Let f : A ! B be a function. Show that if there is a function g : B ! A such that g � f D IA and f � g D IB;

then f is a bijection. (Comment: Such a g is f �1:)(ii) Show that if f : A ! B and h : B ! C are bijections, then h � f : A ! C is a bijection.

16. Let A; B be subsets of Rand f : A ! B be a function. If for every b 2 B; the horizontal line y D b intersectsthe graph of f at most once, then show that f is injective. If ‘at most once’ is replaced by ‘at least once’, whatcan be said about f ?

For each of the sets in exercises 17 to 26, determine if it is countable or uncountable:

17. intervals .a; b/ and [a; b], where we assume a < bI18. Q� .RnQ/I19. A D f 1

2nC 1

3m: n;m 2ZgI

20. B D fx Cp2y : x; y 2 NgI

21. the set C of all lines inR2 passing through the origin;

22. D D fx 2 R : x5 C x C 2 2 QgI23. the set E of all circles inR2 with centers at rational coordinate points and positive rational radius.

24. the set F D fa : x4 C ax � 5 D 0 has a rational rootgI25. the set G D fa3 C b3 : a 2 X; b 2 Y g; where X is a nonempty countable subet of R and Y is an uncountable

subet ofR;

26. the set H D .X n Y /[.Y n X /; where X is a countable set and Y is an uncountable set. (Remark: The set.X n Y /[.Y n X / is called the symmetric difference of X and Y and is usually denoted by X4Y: This conceptwill appear in other algebra and analysis courses later.)

27. Show that the set F of all finite subsets of N is countable.

28. If S is a countable subset of R2; show that for any two points x; y 2 R2 n S; there is a parallelogram in R2 n Shaving x; y as opposite vertices. Here parallelogram means only the 4 edges (including the 4 vertices, but notincluding any interior point).

61

*29. From K0 D [0; 1]; remove the middle thirds to get K1 D [0; 1] n . 13 ; 2

3 / D [0; 13 ][ [ 2

3 ; 1]: Then remove the middlethirds of the 2 subintervals of K1 to get K2 D [0; 1

9 ] [ [ 29 ; 1

3 ] [ [ 23 ; 7

9 ] [ [ 89 ; 1]: Inductively, remove the middle

thirds of the 2n subintervals of Kn to get KnC1: The set K D K0 \ K1 \ K2 \ K3 \ � � � is called the Cantor set.Prove that K is uncountable.(Hint: Consider base 3 representations, i.e. representations of the form.:a1a2a3 : : :/3 D a1

3C a2

32C a3

33C � � � ;

where each ai D 0; 1 or 2. What do the base 3 representations of numbers in Kn have in common? Note somenumbers have 2 representations, e.g. 1

3 D .:1000 : : :/3 and .:0222 : : :/3:)30. For each of the following series, determine if it converges or diverges.

(a)16

kD1cos

�sin

1

k

�(term test) (b)

16kD1

1pk.k C 1/.k C 2/

(c)16

kD1ln�

1C 1k

�(find sum) (d)

16kD1

�12C 1

k

�k

(e)16

kD2

ln k

k(f)

16kD1

cos 2k

k2

(g)16

kD1

k C 2

k C 1

�2

3

�k

(h)16

kD1

cos k�pk

(i)16

kD1ke�k2

(j)16

kD1

k.k C 1/!(k)

16kD1

Arctan k

k2 C 1(l)

16kD1

1

k1C 1k

(compare with a p-series)

(m)16

kD1tank

�k C 1

k

�(n)

16kD1

�1� cos

1

k

�(compare with a p-series)

(o)16

kD1k2 sinp

�1k

�(depends on p) (p)

16kD1

.�1/kC1.pk C 1�pk/

31. Let a1 � a2 � a3 � : : : � 0: Prove that16

kD1ak converges if and only if

16kD1

2ka2k converges. (This is called

Cauchy’s condensation test.) Use this test to determine if16

kD3

1

k ln k ln.ln k/ converges.

32. Show that16

kD2

k

2k�1D 2

2C 3

22C 4

23C � � � converges and find the sum. (Hint: Compare the same series with

index from 1 to 1:)(Probabilistic interpretation: the sum is the expected number of births to get babies of both sex. The probability

of the k-th birth finally resulted in babies of both sex is22kD 1

2k�1:)

*33. Let pn be the n-th prime number, i.e. p1 D 2; p2 D 3; p3 D 5; p4 D 7; � � � : Show that16

kD1

1pk

diverges. (Hint:

Suppose it converges to s:Then the partial sum sn has limit s; as n !1:So for some n; s�sn D 16kDnC1

1pk

< 12:Let

Q D p1 p2 � � � pn : Show, by considering the prime factorization of 1CmQ; thatN6

mD1

1

1CmQ� 16

jD1

� 16kDnC1

1

pk

� j

for every positive integer N ; which will lead to a contradiction.)

62

For exercises 34 to 36, use definitions, Archimedean principle, density of rationals, density of irrationals, etc. tosupport your reasonings.

34. For each of the following sets, if it is bounded above, give an upper bound and find its supremum with proof. Ifit is bounded below, give a lower bound and find its infimum with proof.

(a) A D fpm Cpn : m; n 2 Ng (b) B D .�1; �] [ �4� 1

n: n 2 N�

(c) C D �1nC 1

2m: m; n 2 N� (d) D D Q\ .0;p2]

35. Let A and B be nonempty subsets ofR; which are bounded above. Let

S D fx y : x 2 A; y 2 Bg and T D fx � y : x 2 A; y 2 Bg:Must S or T be bounded above? Give a proof if you think the answer is ‘yes’ or give a counterexample if youthink the answer is ‘no’.

36. Let A and B be nonempty subsets ofR; which are bounded above. Let

C D fx C y : x 2 A; y 2 Bg:Show that C is bounded above and sup C D sup A C sup B: (Hint: If sup C < sup A C sup B; then consider" D .sup AC sup B � sup C/=2: Apply supremum property to get a contradiction.)

37. Let wn D 4n C 5

n3; then fwng should converge to 0. For a given " > 0; show there is a positive integer K such

that if n � K ; then jwn � 0j < ": If " D 0:1; give one such positive integer K :38. Let x be positive and an D [x]C [2x]C � � � C [nx]

n2: Show that fang converges to x

2 by the squeeze limit theorem.

(Here [y] is the greatest integer less than or equal to y:)39. Show that if x is a real number, then there is a sequence of rational numbers converging to x :40. If lim

n!1 an D A; then show that limn!1 janj D jAj: (Hint: Show

��jxj � jyj�� � jx � yj for x; y 2 R first.) Is the

converse true?

41. If fang converges to A; then�

an C anC1

2

�should converge to

AC A

2D A: Prove this by checking the definition.

42. Let x1 D 4 and xnC1 D 4.1C xn/4C xn

for n D 1; 2; 3; : : : : Plot the first 3 terms on the real line. Then prove the

sequence fxng converges.

43. Show that the sequencen�

1C 1

n

�nois increasing and bounded above.

44. Let fxng be a bounded sequence inR; Mn D supfxn; xnC1; xnC2; : : :g and mn D inffxn; xnC1; xnC2; : : :g for n 2 N:(a) Prove that both sequence fMng and fmng converge. (The limit of Mn is called the limit superior of xn and isdenoted by limsup

n!1 xn; while the limit of mn is called the limit inferior of xn and is denoted by liminfn!1 xn:)

(b) Prove that limn!1 xn D x if and only if lim

n!1 Mn D x D limn!1 mn (i.e. limsup

n!1 xn D x D liminfn!1 xn).

63

45. Let x1 D 1; x2 D 2 and xnC1 D xn C xn�1

2for n D 2; 3; 4; : : : : Plot the first 4 terms on the real line. Then prove

the sequence fxng converges.

46. If16

kD2jxk � xk�1j converges, then show the sequence fxng is a Cauchy sequence by checking the definition of

Cauchy sequence.

47. If an � 0 for all n 2 N and fang is a Cauchy sequence, then show that fpang is also a Cauchy sequence bychecking the definition of Cauchy sequence.

48. Let 0 < k < 1: If jxnC1 � xnj � kjxn � xn�1j for n D 2; 3; 4; : : : ; then prove that fxng is a Cauchy sequence.

49. Prove that if fang converges to A; then f�ng converges to A; where �n D a1 C a2 C � � � C an

n: Show that the

converse is false.

(Hint: Let bn D an � A and �n D b1 C b2 C � � � C bn

n; then the first part of the problem becomes showing f�ng

converges to 0. Here use fbng converges to 0 and so jbK0j; jbK0C1j; : : : is small when K0 is large. For n � K0;write �n D b1 C b2 C � � � C bK0�1

nC bK0 C � � � C bn

n:/

50. If fxng is bounded and all its convergent subsequences have the same limit x; then prove that limn!1 xn D x :

51. Let S D f2�n : n 2 Ng and f : N! S is injective. Show that limn!1 f .n/ D 0:

*52. Let fang be a sequence satisfying limn!1.anC1 � an

2/ D 0: Prove that lim

n!1 an D 0:*53. Let fxng be a sequence satisfying lim

n!1.xn � xn�2/ D 0: Prove that limn!1 xn � xn�1

nD 0:

*54. Let fxng be a sequence and let y1 D 0 and yn D xn�1 C 2xn for n D 2; 3; 4; : : : : If fyng converges, prove that fxngalso converges.

55. For a sequence fang of nonzero numbers, we say the infinite product1Y

nD1

an converges to a nonzero number L (or

has value L) if limk!1 a1a2 � � �ak D L : We say it diverges if the limit is 0 or does not exist. Determine if each of the

following infinite products converges or diverges. Find the values of the infinite products that converge.

(a)1Y

nD2

�1� 2

n.n C 1/� (b)1Y

nD2

�1� 1

n2

�(c)

1YnD2

n3 � 1

n3 C 1(d)

1YnD0

.1C z2n / for jzj < 1:Remarks: In Apostol’s book, the following theorem is proved: In the case every an � 0; we have

1YnD1

.1 C an/;1YnD1

.1 � an/; 16nD1

an all converge or all diverge. Try to do the above exercises without using this theorem.

56. Prove that every bounded infinite subset of Rhas an accumulation point. (This is also often called the Bolzano-Weierstrass theorem.)

64

57. Let f : .0;C1/! Rbe defined by f .x/ D x

x C 1: Show that lim

x!1f .x/ D 1

2by checking the definition.

58. Define f : R! Rby

f .x/ D n8x if x is rational,2x2 C 8 if x is irrational.

For which x0; does limx!x0

f .x/ exist? (Hint: Sequential limit theorem.)

59. If f : R! R is continuous and f .x/ D 0 for every rational number x; show that f .x/ D 0 for all x :60. (a) Find all functions f : Q! Rsuch that f .x C y/ D f .x/ C f .y/ for all x; y 2 Q:

(b) Find all strictly increasing functions f : R! R such that f .x C y/ D f .x/ C f .y/ for all x; y 2 R:61. For a function f : R! R; we say f has a local (or relative) maximum at x0 if there exists an open interval.a; b/ containing x0 such that f .x/ � f .x0/ for every x 2 .a; b/: Similarly, we say f has a local (or relative)

minimum at x1 if there exists an open interval .c; d/ such that f .x/ � f .x1/ for every x 2 .c; d/: If f : R! Riscontinuous and has a local maximum or a local minimum at every real number, show that f is a constant function.

62. If f .x/ D x3; then f . f .x// D x9: Is there a continuous function g : [�1; 1] ! [�1; 1] such that g.g.x// D �x9

for all x 2 [�1; 1]? (Hint: If such a function g exists, then it is injective.)

63. A fixed point of a function f is a number w such that f .w/ D w: Show that if f : [0; 1] ! [0; 1] is continuous,then f has at least one fixed point. (Hint: Consider g.x/ D f .x/ � x :)

64. Let f : [0; 1] ! [0; 1] be an increasing function (perhaps discontinuous). Suppose 0 < f .0/ and f .1/ < 1;show that f has at least one fixed point. (Hint: Sketch the graph of f and consider the set ft 2 [0; 1] : t < f .t/g.Does it have a supremum?)

65. Let f : R! R be a continuous function such that j f .x/ � f .y/j � jx � yj for all x; y 2 R: Show that f isbijective. (Hint: Easy to show f is injective. To show f is surjective, let w 2 Rand M D jw� f .0/j, show thatw and f .0/ are between f .�M/ and f .M/:)

*66. Let f : [0; 1] ! .0;C1/ be continuous and M D supf f .x/ : x 2 [0; 1]g: Show that

limn!1�Z 1

0f .x/ndx

� 1n D M

if the limit exists. (Hint: M D f .x0/: For every k 2 N; use the sign preserving property to show thatf .x0/ � 1

k � f .x/ � M on an interval containing x0. Squeeze the integral.)

67. Find the derivatives of the functions f .x/ D �x2 if x 6D 0x if x D 0

and g.x/ D j cos xj:68. Let f : R! Rbe differentiable at c and In D [an; bn] be such that I1 � I2 � I3 � � � � and

1\nD1

[an; bn] D fcg:Prove that if an < bn for all n 2 N; then f 0.c/ D lim

n!1 f .bn/� f .an/bn � an

:69. Let f .x/ D jxj3 for every x 2 R: Show that f 2 C2.R/: However, f 000.0/ does not exist.

70. Let f : R! R satisfies j f .a/ � f .b/j � ja � bj2 for every a; b 2 R: Show that f is a constant function. Ifthe exponent 2 in the inequality is replaced by a number greater than 1, must f be a constant function? If 2 isreplaced by 1, must f be a constant function?

65

71. Let n be a positive integer and f .x/ D .x2 � 1/n : Show that the n-th derivative of f has n distinct roots.

72. Let f : [0;1/! Rbe continuous and f .0/ D 0: If f 0.x/ � f .x/ for every x > 0; then show that f .x/ � 0 forevery x 2 [0;1/: (Hint: What if f 0.x/ D f .x/?)

73. If f : .0;C1/ ! R is differentiable and j f 0.x/j � 2 for all x > 0; then show that the sequence xn D f�1

n

�converges. Also, show lim

x!0C f .x/ exists. (Hint: Check fxng is a Cauchy sequence. For the second part, consider

the sequential limit theorem and the remarks following it.)

74. For 0 < x < �2; prove that j ln.cos x/j < x tan x :

75. Let f : [0;1/ ! Rbe continuous and f .0/ D 0: If j f 0.x/j � j f .x/j for every x > 0; show that f .x/ D 0 forevery x 2 [0;1/: (Hint: Let j f j has maximum value M on [0; 1

2 ]: Apply the mean value theorem to f on [0; 12 ]:)

76. Let f : R! Rbe differentiable. If f 0 is differentiable at x0; show that

limh!0

f .x0 C h/C f .x0 � h/ � 2 f .x0/h2

D f 00.x0/:77. Prove that if 0 � � � �

2 , then

1� �2

2� cos � � 1� �2

2C �4

24:

(Hint: Apply Taylor’s theorem to the four times differentiable function cos �:)78. Let f : .0;C1/ ! Rbe twice differentiable, M0 D supfj f .x/j : x > 0g <1; M1 D supfj f 0.x/j : x > 0g <1

and M2 D supfj f 00.x/j : x > 0g < 1: Show that M21 � 4M0M2: (Hint: Let h > 0: Apply Taylor’s theorem to

f .x/ with x D cC 2h; then solve for f 0.c/:)79. (a) If f : .a; b/! Ris differentiable and j f 0.x/j � 2 for all x 2 .a; b/; then show that f is uniformlycontinuous.

(b) Show that f : .0;C1/ ! Rdefined by f .x/ D sin1

xis not uniformly continuous.

80. (a) Prove that if the union of a collection of open intervals contains [a; b]; then there are finitely many of theseintervals, whose union also contains [a; b]: (Hint: Suppose this is false. Let m1 D .aCb/=2: Then one of [a;m1]or [m1; b] is not contained in the union of finitely many of these open intervals. Proceed as in the proof of theBolzano Weierstrass theorem.)

(b) Give a proof of the uniform continuity theorem using part (a).

81. If f is continuous on [a; b]; f .x/ � 0 for all x 2 [a; b] andZ b

af .x/ dx D 0; then prove that f .x/ D 0 for all

x 2 [a; b]:82. Let f : [a; d] be a bounded function and a � b � c � d:

(i) Use the integral criterion to show that if f is integrable on [a; b] and [b; c]; then f is integrable on [a; c]:(ii) Use the integral criterion to show that if f is integrable on [a; d]; then f is integrable on [b; c]:

83. Let f : [a; b] ! Rbe bounded and fx 2 [a; b] : f is discontinuous at xg D fx1; x2; : : : ; xng; where x1 < x2 <� � � < xn : Use the integral criterion to show that f is integrable on [a; b]: (Hint: Divide [a; b] into subintervalswhere f is continuous except at one endpoint and note part (i) of problem above.)

66

84. (i) Let f; g : [a; b] ! Rbe bounded and P is a partition of [a; b]: Show that

L. f; P/ C L.g; P/ � L. f C g; P/ � U . f C g; P/ � U . f; P/ CU .g; P/:(ii) If f and g are integrable on [a; b]; show thatZ b

a. f .x/ C g.x// dx D Z b

af .x/ dx C Z b

ag.x/ dx :

(Hint: First show RHS� " < L. f; P/ C L.g; P/ � L. f C g; P/ for some P using supremum property.)

85. Determine if each of the following improper integrals exists or not.

(a)Z 1

0

dxpex

(b)Z 1

0sin x dx

(c)Z 1

0

dx

x2 C 5x(d)

Z 1�1

dx3p

x

(e)Z 1

0

dx

x.x � 1/ (f)Z 1

0

cos x

1C x2dx

86. Find the principal value of each of the following integrals if it exists.

(a) P.V.Z C1�1 x

ex2 dx (b) P.V.Z 2

0

dx

x2 � 1

87. Prove that for 0 < x < 1; the improper integral 0.x/ D Z 10

t x�1e�t dt exists. This is called the gamma

function. (Hint: Consider the cases x 2 .0; 1/ and x 2 [1;1/ separately.)

Past Exam Problems

88. Define the following terms:(a) S is a countably infinite set(b) S is a countable set

(c) a series16

kD1ak converges to a number S

(d) a nonempty subset S ofR is bounded above(e) the supremum of a subset S ofR that is bounded above(f) a sequence fxng converges to a number x(g) a sequence fxng is a Cauchy sequence(h) x is an accumulation point of a set S(i) f : S ! Rhas a limit L at x0

(j) f : S ! R is continuous at x0 2 S (the "-� definition)

89. For each of the following sets S, determine if it is countable or uncountable. Be sure to give reasons to supportyour answer.(a) S is the set of all intersection points .x; y/ of the line y D �x with the graphs of all equations y D x3CxCm;

where m 2Z:(b) S is the set of all intersection points .x; y/ of the graph of y D x3 C x C 1 with all lines y D mx; where

m 2Z:(c) S is the set of all intersection points .x; y/ of the circle x2 C y2 D 1 with all hyperbolas x y D 1

m; where

m 2 N:67

(d) S D fa C b 2 R : jaj 2 M; b 2 Qg; where M is a uncountable subset of the positive real numbers.(e) S D fa C b 2 R : jaj 2 M; b 2 Qg; where M is a countable subset of the positive real numbers.

(f) S D Q.p2/ D na C bp

2

cC dp

2: a; b; c; d 2 Q; cC d

p2 6D 0

o:(g) S D fx2 C y2 C z2 : x 2 A \ B; y 2 Q\ A; z 2 B \Qg; where A is a nonempty countable subset ofRand B is an uncountable subset ofR:(h) S D fx � y : x; y 2 Ag; where A is an uncountable subset ofR:(i) S D fx2 C y2 : x; y 2 Ag; where A is a nonempty countable subset ofR:(j) S D fx C sin y : x 2 Rn A; y 2Zg; where A is a nonempty countable subset ofR:(k) S D f.x; y/ 2 R2 : x 2 A; y 2 Rn Ag; where A is a nonempty countable subset ofR:(l) S D fx C y

p2 : x 2Z; y 2 Ag; where A is a nonempty countable subset ofR:

(m) S D Rn fa C bp

2� cp

3 : a; b; c 2 T g; where T D fr� : r 2 Qg:(n) S D T \U; where T D RnQ and U D Rn fpm Cpn : m; n 2 Ng: (Hint: ConsiderRn .T \U /:)(o) S is the set of all squares on the plane that can be circumscribed by circles with rational radii and centers

with rational coordinates.(p) S is the set of all nonconstant polynomials with coefficients in G; where i D p�1 and G D faCbi : a; b 2Zg:

90. Determine if each of the following series converges or diverges. Be sure to give reasons to support your answer.

(a)16

kD1

cos k�k2 C 2k

and16

kD1

ep

kpk

(b)16

kD1

.2k/!3kk4

and16

kD1

.cos k/.sin 2k/2k

(c)16

kD1

12

�cos

1kC sin

1k

�and

16kD1

sin�1

k� 1

k C 1

�(d)

16kD1

2k C 3k

1k C 4kand

16kD1

cos�

sin1k

�(e)

16kD1

21=k C 31=k

11=k C 41=kand

16kD1

.cos k�/�sin1

k� �(f)

16kD1

.k!/2.k2/! and16

kD1

�cos

1

k

��sin

1

k

��tan

1

k

�(g)

16kD2

2k cos k.k � 1/! and16

kD2

sin. 1k /

ln k

(h)16

kD1

k� C cos k�3C k4

and16

kD1

k� cos k�3k4

(i)16

kD2

.2k/!.k C 1/!.k � 1/! and16

kD1k cos

�1

k2

�(j)

16kD1

.3k/!k!.2k/! and

16kD2

cos.1=k/k2 � 1

(k)16

kD1

k!.2k � 1/! and16

kD1

cos k�pk C 1

(l)16

kD1

2kk2

k!and

16kD1

1pk

sin

�1pk

�(m)

16kD1

1k cos k� and

16kD1

k2 sin.1=k/.2k C 1/!(n)

16kD1

cosk.1C 1

k/ and

16kD1

cos.sin.1=k//sin.cos.1=k//

91. Determine if each of the following set has an infimum and a supremum. If they exists, find them and give reasonsto support your answers.

(a) S D n 1

mC 1

n: m; n 2 No n n2

k: k 2 No

68

(b) S D nx C y : x; y 2 �1

2; 1�o n n2� 1

n: n 2 No

(c) S D nx � 1

n: x 2 [0; 1]\Q; n 2 No n ��1; 1

2

�(d) S D n x � �

x C � : x 2 .RnQ/ \ [�;1/o(e) S D n x � �

x C � : x 2 Q\ [0;1/o(f) S D fx3 C y3 : x 2 Q\ [0; 1]; y 2 .�1; 0]g(g) S D n p

2

m C nC 1

kp

2: m; n; k 2 No

(h) S D 1[kD1

[1� 1

2k � 1; 1� 1

2k/

(i) S D npx C y2 : x; y 2 .0; 1] \Qo

(j) S D n1

nC x : x 2 [0; 1]\Q; n D 1; 2; 3; : : :o

(k) S D fx C y : x 2 [0; 1]\Q; y 2 [0; 1]\ .RnQ/g(l) S D fx 2 R : x.x C 1/ � 0 and x 2 RnQg

(m) S D f kn! : k; n 2 N; k

n! < p2g

(n) S D 10[nD1

�� 1

np

2; 2� 1

n

� nQ�(o) S D fx2 C y3 C z4 : x 2 .�1; 0/ nQ; y 2 .0; 1/ \Q; z 2 .�1; 1/g

92. For each of the following sequences fxng; show it converges and find its limit.

(a) x1 D 1 and xnC1 D xn

2Cp

xn for n D 1; 2; 3; : : : :(b) x1 D 1; x2 D 2 and xnC1 D p

xn Cpxn�1 for n D 2; 3; 4; � � �:(c) x1 D 1 and xnC1 D 2� xn

3C xnfor n D 1; 2; 3; : : : :

(d) x1 D 1516 and xnC1 D 1�p1� xn for n D 1; 2; 3; � � � : Also, do the sequence

n xnC1

xn

o:(e) xn D anC1

an; for n D 1; 2; 3; : : : ; where a1 D 1; a2 D 2 and anC1 D an C an�1 for n D 2; 3; 4; � � � :

(f) x1 D 1 and xnC1 D 1� 1

4xnfor n D 1; 2; 3; : : : :

(g) x1 D 4 and xnC1 D 12

�xn C 4

xn

�for n D 1; 2; 3; : : : : (Hint: Sketch the graph of f .x/ D x C 4

xfor x � 2:)

(h) x1 D 5 and xnC1 D 3C 4xn

for n D 1; 2; 3; � � � :(i) x1 D 2 and xnC1 D 2� 1

xnfor n D 1; 2; 3; � � � :

(j) x1 D 0 and xnC1 D x2n C 4

5for n D 1; 2; 3; : : : :

(k) x1 D 1 and xnC1 D pxn � 1

4for n D 1; 2; 3; : : : :

(l) x1 D 3 and xnC1 D s1� 1

xn C 1for n D 1; 2; 3; : : : :

(m) 0 < x1 < 1 and 7xnC1 D x3n C 6 for n D 1; 2; 3; : : : : (Hint: x3 � 7x C 6 D .x � 1/.x � 2/.x C 3/:)

(n) x1 2 [1;1/ and xnC1 D p3xn � 2 for n D 1; 2; 3; : : : :

(o) x1 D 0; x2 2 [0; 12 ] and xnC1 D 1

3 .1C xn C x3n�1/ for n D 2; 3; 4; : : : :

93. If fang is a decreasing sequence (of nonnegative real numbers) converging to 0, show that the sequence fxng69

converges, where

xn D n6kD1

.�1/kC1ak D a1 � a2 C a3 � � � � C .�1/nC1an :94. Let 0 < a < b and a1 D a; b1 D b; anC1 D an C bn

2; bnC1 D ra2

n C b2n

2for n D 1; 2; 3 : : : : Show that fang

and fbng both converge and their limits are the same.

95. (i) If a � b and 0 < t < 1, then show that a � ta C .1 � t/b � b:(ii) Let

x1 D 1; x2 D 2 and xnC1 D 13

xn C 23

xn�1 for n D 2; 3; 4; � � � :Show that the sequence fxng converges.

96. Show that the sequence fxng given by

x0 D 0; x1 D 1 and xnC1 D r1

4x2

n C 3

4x2

n�1 for n 2 Nconverges and find its limit.

97. Let S be a set of real numbers such that every sequence in S has a convergent subsequence, show that S is bounded.

98. Let A be a nonempty subset of the nonnegative real numbers. If A is bounded above and B D fx2C y2 : x; y 2 Ag;show B is bounded above and sup B D 2.sup A/2 :

99. Suppose An � .�1; 2/ and xn D sup An for n D 1; 2; 3; : : : ; 10: Prove that

sup� 10[

kD1

An� D max.x1; x2; : : : ; x10/:

100. Use the definitions of infimum and supremum to explain the statement: Let f : R�R! [0; 1] be a function,g.x/ D supf f .x; y/ : y 2 Rg and h.y/ D inff f .x; y/ : x 2 Rg: Show that supfh.y/ : y 2 Rg � inffg.x/ : x 2Rg:

101. Show that for every x 2 R; there is a strictly increasing sequence of irrational numbers fxng converging to x :102. Prove that lim

n!1� 1

n2� p

2

n3

� D 0 by checking the definition of limit.

103. Prove that limn!1� 2

n C 1� 1

n2

� D 0 by checking the definition of limit.

104. Given limn!1 xn D 0: Prove that lim

n!1�xn C 1n

� D 0 by checking the definition of limit.

105. Given limn!1 xn D 1

2: Prove that lim

n!1 xnn D 0 by checking the definition of limit.

106. Given limn!1 xn D 8: Prove that lim

n!1 3p

xn D 2 by checking the definition of limit.

70

107. Given sequences fxng and fyng both converge to A: For n 2 N; let zn D max.xn; yn/: Show that fzng convergesto A by checking the definition.

108. If fxng is a sequence such that jxkC1 � xkj < 1

2kfor k D 1; 2; 3; : : : ; then show that fxng is a Cauchy sequence.

109. (a) Let S be an open interval and x0 2 S: For a function f : S ! R; state the definition of f .x/ converges to L(or has limit L) as x tends to x0 in S:(b) Let f : .1; 3/! Rbe defined by f .x/ D x2 C 1

x: Prove that lim

x!2f .x/ D 9

2by checking the definition.

(c) Let f : .1; 4/! Rbe defined by f .x/ D jx2 � 9j: Prove that limx!2

f .x/ D 5 by checking the definition.

110. If f; g : .0; 1/! R are increasing functions, show that h.x/ D max. f .x/; g.x// has countably many jumps onthe interval .0; 1/:

111. Give an example of a function f : R! R that is continuous at x 2Z; but discontinuous at every x 62Z: Be sureto give reasons to support your example.

112. (a) State the intermediate value theorem.(b) Let f : [0; 2] ! Rbe continuous and f .0/ D f .2/: Show that there exists c 2 [0; 1] such that f .c/ D f .cC1/:(c) Show that there is a nonzero continuous function g : R! Rsuch that g.t/C g.2t/C g.3t/ D g.4t/C g.5t/for every t 2 R: (Hint: Try g.t/ D jt jr for some constant r:)

113. (a) Show that the set T D fx : sin x 2 Qg is countable. (Hint: For a fixed rational r; how many solutions ofsin x D r are there in the interval [k�; k� C 2�/?)(b) If f : [0; 1] ! Ris continuous and sin f .x/ 2 Q for every x 2 [0; 1]; then show that f is a constant function.

114. Show that there is no continuous function f : R! Rsuch that for every c 2 R, f .x/ D c has exactly 2 solutions.

115. Suppose f : R! R is a function such that f .x C y/ D f .x/C f .y/ for every x; y 2 Rand j f .x/j � x4=jxj forevery x 6D 0:(a) Show that f is continuous at some x 2 R:(b) Show that f is continuous at every x 2 R:(c) Give an example of such a function.

116. Suppose f; g : [1; 2] ! [3; 4] are continuous functions and also fg.x/ : x 2 [1; 2]g D [3; 4]: Show that there isc 2 [1; 2] such that f .c/ D g.c/:

117. Let f : R! Rbe continuous such that j f .x/ � f .y/j � 12 jx � yj for every x; y 2 R:

(a) Let w 2 R: Define x1 D w and xnC1 D f .xn/ for n 2 N: Show that fxng is a Cauchy sequence. (Hint: How isjxkC1 � xk j compared to jx2 � x1j?)(b) Show that there is x 2 R such that f .x/ D x :

118. Give an example of a function f : .0; 2/ ! R that is differentiable for every x 2 .0; 2/; but f 0.x/ is notcontinuous at x D 1: Be sure to give reasons to support your example.

119. Let f : .0; �/ ! R satisfiespj f .a/ � f .b/j � sin ja � bj for every a; b 2 .0; �/: Show that f is a constant

function.

120. (a) State the mean value theorem.(b) For the function f .x/ D sin x; find the smallest constant K such that j f .b/ � f .a/j � K jb � aj holds forevery a; b 2 R:

121. If f : R! R is differentiable and limx!0

f 0.x/ exists, then show that f 0 is continuous at 0.

71

122. Use the mean value theorem to prove that f .x/ D sin 5x is uniformly continuous.

123. Prove that if f : R! .0;C1/ is uniformly continuous, then the function g.x/ D pf .x/ is also uniformly

continuous.

124. (a) State Lebesgue’s theorem.(b) Let f; g : [0; 2] ! [0; 1] be Riemann integrable. Prove that the function h : [0; 2] ! [0; 1] defined by

h.x/ D �f .x/ if x 2 [0; 1/g.x/ if x 2 [1; 2]

is Riemann integrable.

125. If f; g : [0; 1] ! R are Riemann integrable, show that the function h.x/ D min. f .x/; g.x// is Riemannintegrable on [0; 1]:

126. For every positive integer n; give an example of a Riemann integrable function fn : [0; 1] ! [0; 1] such thatlim

n!C1 fn.x/ exists for every x 2 [0; 1]; but the function f .x/ D limn!C1 fn.x/ is not Riemann integrable on [0; 1]:

Be sure to give reasons to support your example.

127. (a) Determine if the improper integralZ 1�1 cos 3x

1C x2dx exists or not.

(b) Determine if the principal value P.V.Z 1�1 cos 3x

1C x2dx exists or not.

(c) Determine if the improper integralZ 1�1

13p

xdx exists or not.

(d) Determine if the principal value P.V.Z 1�1

13p

xdx exists or not.

(e) Determine if the improper integralZ 1�1 sin x dx exists or not.

(f) Determine if the principal value P.V.Z 1�1 sin x dx exists or not.

128. (2002 L1 Midterm) Let f .x/ D x2 C 3: Determine if the set S D f f .w C z/ : w 2 [0; 1]\Q; z 2 [2; 3] nQg hasan infimum and a supremum. If they exist, find them and give reasons to support your answers.

129. (2002 L1 Midterm) Determine if each of the series16

kD1

2kp

k.2k/! and16

kD1.cos k/�sin

1k2

�converges or diverges.

130. (2002 L1 Midterm) Let S be the set of all lines L on the coordinate plane such that L passes through two distinctpoints in Q�Q and T be the set of all points, each of which is the intersection of a pair of distinct lines in S:Determine if T is a countable set or not.

131. (2002 L1 Midterm) Given xn 6D �1 for all n 2 N: If limn!1 xn D 0; then show that lim

n!1 xn

1C xnD 0 by checking

the definition of limit.

132. (2002 L1 Midterm) Given fxng converges to w 2 Rand xn < w for all n 2 N: For every n 2 N; let

yn D supn

x2k : k 2 N; k � n C 1

2

o:Show that fyng converges to w:

133. (2002 L2 Midterm) Let f .x/ D sin x : Determine if the set S D nf .w/ � 1

n: w 2 ��

4; �

3

� nQ; n 2 No has an

infimum and a supremum. If they exist, find them and give reasons to support your answers.

72

134. (2002 L2 Midterm) Determine if each of the series16

kD1

.2k C 1/5

k!and

16kD1

cos k

k4 C k C 1converges or diverges.

135. (2002 L2 Midterm) Let S be the set of all circles on the coordinate plane that pass through .1; 1/ and anotherpoint .xp2; x

p2/ for some x 2 Q: Determine if S is a countable set or not.

136. (2002 L2 Midterm) Given xn 6D 1 for all n 2 N: If limn!1 xn D 0; then show that lim

n!1 xn

xn � 1D 0 by checking the

definition of limit.

137. (2002 L2 Midterm) Given fxng converges to w 2 Rand xn > w for all n 2 N: For every n 2 N; let

yn D infn

x2k : k 2 N; k � n C 1

2

o:Show that fyng converges to w:

138. (2002 L3 Midterm) Let f .x; y/ D x2 C y2: Determine if S D �f�1C .�1/n ; w� : n 2 N; w 2 [1;p2/ nQ has

an infimum and a supremum. If they exist, find them and give reasons to support your answers.

139. (2002 L3 Midterm) Determine if each of the series16

kD1

2k

k3.3k/! and16

kD1

�1eC 1

k

�ksin k converges or diverges.

140. (2002 L3 Midterm) Let x1 D 1 and xnC1 D 12

qx2

n C 4xn for n D 1; 2; 3; : : : : Show that fxng converges. Find the

limit of fxng:141. (2002 L3 Midterm) Let S be the set of all ordered pairs .p;C/; where p D .x; y/ 2 Q�Q and C is the circle

with center p and radius jx yj C 1: Determine if S is a countable set or not.

142. (2002 L3 Midterm) If limn!1 xn D 1; then show that lim

n!1.x2n � 1/ D 0 by checking the definition of limit.

143. (2002 L3 Midterm) Let fxng be a bounded sequence of real numbers. Let

S D fw : there exist infinitely many n 2 N such that w < xng:(a) Show that S is nonempty and bounded above.

(b) Show that fxng has at least one subsequence fxnk g converging to s D sup S:144. (2004 L1 Midterm) Find (with proof) all positive numbers b such that the series

16kD1

�b C 1

k

�kconverges.

145. (2004 L1 Midterm) Let A be a nonempty countable subset ofR: Let

S D f� : � 2 R; sin � 2 Ag and T D f� : � 2 R; sin � 62 Ag:Determine (with proof) if each of the sets S and T is countable or uncountable.

146. (2004 L1 Midterm) Show that the sequence fxng given by

x1 D 6 and xnC1 D x2n C 4

2xn C 3for n D 1; 2; 3; : : :

converges and find its limit.

73

147. (2004 L1 Midterm) If xn 6D �1 for all n and limn!1 xn

xn C 1D 1

2; then prove that lim

n!1 xn D 1 by checking the

definition of limit.

148. (2004 L1 Midterm) Let S � [0; �2

]; T D fcos2 a C cos2 b : a; b 2 Sg and U D fsin c : c 2 Sg: If sup T D 1

2;

then find the infimum of U with proof.

149. (2004 L2 Midterm) Let S be the set of all intersection points .x; y/ 2 R2 of the graphs of the equationsx2 Cmy2 D 1 and mx2 C y2 D 1; where m 2Zn f�1; 1g: Determine if S is countable or uncountable. Provide aproof of your answer.

150. (2004 L2 Midterm) Let x1 D 9 and xnC1 D pxn C 2xn

3for n D 1; 2; 3; : : : : Prove that fxng converges. Find the

limit of fxng:151. (2004 L2 Midterm) Let ak > 0 for k D 1; 2; 3; : : : and

16kD1

ak converges. Determine all positive real number b

such that the series16

kD1

.b C ak/k

kconverges. Be sure to give a proof of your answers.

152. (2004 L2 Midterm) Prove that if limk!1 x2k D 0:5 and lim

k!1 x2kC1 D 0:6; then the sequence limn!1 xn

n D 0 by checking

the definition of limit.

153. (2004 L2 Midterm) Let I be a nonempty set. For every t 2 I; let At be a nonempty subset of [0; 1]: Letxt D sup At : Prove that if A D[

t2I

At ; then sup A D supfxt : t 2 I g:154. (2002 Final) Determine if the improper integral

Z 1

0

cos 3xpx

dx exists or not.

155. (2002 Final) Prove that there does not exist any continuous function f : R! R such that f�

f .x/� C x D 0 forevery x 2 R:

156. (2002 Final) Let fxng be a Cauchy sequence of real numbers. Prove that fsin 5xng is also a Cauchy sequence bychecking the definition of a Cauchy sequence.

157. (2002 Final) Let f; g : [0; 2] ! Rbe Riemann integrable. Prove that h : [0; 2] ! Rdefined by

h.x/ D �max

�f .x/; g.x/� if x 2 [0; 1]

min�

f .x/; g.x/� if x 2 .1; 2]

is also Riemann integrable on [0; 2]:158. (2003 Final) Determine (with proof) if each of

16kD1

sink�1C 1k

�and

16kD1

1� cos.1=k/1=k2

converges or not.

159. (2003 Final) Let A be a nonempty subset of R such that inf A D 0 and sup A D 1: Determine the infimum andsupremum of S D fa3 � 4a C 1 : a 2 Ag:

160. (2003 Final) Let P be a countable set of points inR2: Prove that there exists a circle C with the origin as centerand positive radius such that every point of the circle C is not in P: (Note points inside the circle do not belongto the circle.)

161. (2003 Final) Let f : R! [u; v] be a function and w 2 R: For every r > 0; let

M.r/ D supf f .t/ : 0 < jt �wj < rg and m.r/ D inff f .t/ : 0 < jt �wj < rg:74

(a) Prove that limr!0C m.r/ and lim

r!0C M.r/ exist.

(b) Prove that limx!w f .x/ D L if and only if lim

r!0Cm.r/ D L D limr!0C M.r/:

162. (2003 Final) Let f : R! Rbe uniformly continuous. Prove that g : R! Rdefined by g.x/ D 1

1C f .x/2is

also uniformly continuous.

163. (2003 Final) Let f : [0; 1] ! [�1; 1] be Riemann integrable. Using the integral criterion, prove that

g.x/ D nf .x/ if 0 < x < 10 if x D 0 or 1

is also Riemann integrable on [0; 1]:164. (2004 Final) Determine whether the improper integral

Z 10

sin x

x3=2dx converge or not.

165. (2004 Final) Let f : R! R be an increasing function and g : R! R be a decreasing function. Prove thatf .x/g.x/ is discontinuous for only countably many x 2 R:

166. (2004 Final) Let f .x/ D(0 if x 2 [0; 1] nQ1 if x D 01=n if x D m=n for m; n 2 Nwith no common prime factor.

Prove that there exists a Riemann integrable function g : [0; 1] ! [0; 1] such that the composition function g � f :[0; 1] ! [0; 1] is not Riemann integrable on [0; 1]:167. (2004 Final) Let f : .0;C1/ ! R satisfy j f .x/ � f .y/j � j sin.x2/ � sin.y2/j for all x; y > 0: Prove that the

sequence x1; x2; x3; : : : given by xn D f .1=n/ is a Cauchy sequence.

168. (2005 Spring Final) Determine the supremum of S D 1[nD1

n1xC 1

np

2: x 2 .2; 3] nQo:

169. (2005 Spring Final) Prove that the function f : R! Rdefined by f .x/ D 2C 3x

x2 C 4is continuous at 2 by checking

the "-� definition of a function continuous at a point.

170. (2005 Spring Final) Let xn > 0 for n D 1; 2; 3; : : : : If fxng is a Cauchy sequence, then prove that fe�xn g is aCauchy sequence by checking the definition of a Cauchy sequence.

171. (2005 Spring Final) (a) Let S � [0; 1]: If S is a set of measure 0, then prove that T D fx2 : x 2 Sg is a set ofmeasure 0.

(b) Let f : [0; 1] ! [0; 1] be integrable. Prove that h : [0; 1] ! [0; 1] defined by h.x/ D f .px/ is integrable.(Caution: In general it is false that f integrable and g continuous on [a; b] imply h D f � g integrable on [a; b]:)

172. (2005 Fall Exam, Version 1) (a) Determine with proof if16

kD1

3k.2k/!k!converges or not.

(b) Let ak 6D �1 for k D 1; 2; 3; : : : and16

kD1jak j converges. Determine with proof if

16kD1

ak

1C akconverges or not.

173. (2005 Fall Exam, Version 1) Define x1 D 4 and xnC1 D 4� 4

xnfor n D 1; 2; 3; : : : : Prove that x1; x2; x3; : : :

converges and find its limit.

174. (2005 Fall Exam, Version 1) Let A be a nonempty bounded subset ofRsuch that sup A D 1 and inf A D 0: Let

S D fx � yp

2 : x 2 A; y 2 [�2; 2/g:75

Determine the supremum and infimum of S with proof.

175. (2005 Fall Exam, Version 1) If limn!1 an D 2; lim

n!1 bn D 3 and all bn 6D �2; then prove that limn!1 an C 3

bn C 2D 1 by

checking the definition of limit.

176. (2005 Fall Exam, Version 2) (a) Determine with proof if16

kD1

.3k/!k!2k

converges or not.

(b) Let cos ak 6D 0 for k D 1; 2; 3; : : : and16

kD1jakj converges. Determine with proof if

16kD1

ak

cos akconverges or not.

177. (2005 Fall Exam, Version 2) Define x1 D 1 and xnC1 D x2n C 15

8for n D 1; 2; 3; : : : : Prove that x1; x2; x3; : : :

converges and find its limit.

178. (2005 Fall Exam, Version 2) Let A be a nonempty bounded subset ofRsuch that sup A D 6 and inf A D 2: Let

S D f x

y� x y : x 2 A; y 2 �1

2; 1�g:

Determine the supremum and infimum of S with proof.

179. (2005 Fall Exam, Version 2) If limn!1 an D 1 and all an 6D n; then prove that lim

n!1 a2n C n

n � anD 1 by checking the

definition of limit.

180. (2006 Spring Exam)(a) Let f : S ! Rbe a function and x0 be an accumulation point of S: State the definitionof f .x/ converges to a real number L as x tends to x0:(b) Let f : .0:5;C1/ ! Rbe defined by f .x/ D r

x C 1

x: Prove that lim

x!1f .x/ D p

2 by checking the definition.

181. (2006 Spring Exam) Define a1 D 1 and anC1 D n

n C 1an C cos n.n C 1/3

for n D 1; 2; 3; : : : : Prove that limn!1 nan

exists inR:182. (2006 Spring Exam) Let a; b 2 Rwith a < b and f : [a; b] ! Rbe continuous. Also, let f .x/ be differentiable

for all x 2 .a; b/: Prove that if the graph of f is not a line segment, then there exist numbers x1 and x2 in the openinterval .a; b/ such that

f 0.x1/ < f .b/ � f .a/b � a

< f 0.x2/:183. (2006 Spring Exam) Let f; g : [0; 1] ! Rbe continuous. If there exists a sequence of numbers x1; x2; x3; : : : 2

[0; 1] such that g.xn/ D f .xnC1/ for n D 1; 2; 3; : : : ; then prove that there exists w 2 [0; 1] such that g.w/ Df .w/:Caution Be careful, xni converges does not imply xniC1 converges !!!

76