1

Chapter 1: Limits of functions:

In this chapter we will cover:

1.1 The tangent and the velocity problems (motivation)

1.2 The limit of a function (intuitive approach)

1.4 The precise definition of the limit of a function

1.3 Limit laws. Limits of elementary functions and the squeeze theorem.

1.6 Limits at infinity and infinite limits . Horizontal and vertical asymptotes.

1.5 Continuity

1.7 Review .

1.1. The tangent and the velocity problems (motivation):

A. The tangent problem:

Consider the graph of a function )(xf , such as the graph shown below (which is the graph of 14)( 3 xxxf ):

Figure 1: Graph of

14)( 3 xxxf

How can we define and find the tangent line to this graph at a point on the curve (say at the point )1,2(P )?

In order to understand how to approach this problem, consider the slightly simpler function: 2)( xxf graphed

below:

Figure 2: Graph of 2)( xxf

2

How can we define and find the equation of the tangent line to this graph at )1,1(P ?

One definition of the tangent line: The tangent line to the graph of a function )(xf at a point ))(,( afaP on the

graph is the line which touches the graph of )(xf only at the point P (if we look only close “enough” to the point)

and which follows the direction of the curve at the point of contact.

How can we obtain a working method from this definition?

Consider two points on the graph of 2)( xxf : )1,1(P and ))(,( xfxQ with 1x .

The points P and Q determine a secant line (a line which intersects the curve more than once) to the graph of )(xf .

Then we determine the slope of the tangent line as PQ

PQmm

lim , where PQm is the slope of the secant line PQ .

In this case (for 2)( xxf ), fill the following table in order to determine the slope m :

Therefore, it appears that 2m in this case, and therefore the equation of the

tangent line to the graph of 2)( xxf at )1,1(P is :

(1) 12121 xyxy

Example :

Following the same method as described above, calculate the equation of the tangent line to the graph of 2)( xxf

at )0,0(P and at )1,1(P .

B. The velocity problem:

As you watch the needle of the speedometer of your car / vehicle, you see that the speed of the car varies in time..

How is this instantaneous speed found at each instant in time ?

Consider that we determine that the position of your vehicle in time is given by the function 29.4)( tts where s(t)

is measured in Kilometers and t is measured in hours. Let us determine the instantaneous speed of this vehicle at

time hour 1t .

Since we do not have (yet) a definition for instantaneous speed, one way to do this is to reme mber that the average

speed of an object over a time interval 21,tt is given by:

12

12, 21

tt

tstsv Avg

tt

(change in position divided

by the change in time over this time interval) and define the instantaneous velocity (or instantaneous speed ) at 1 as

x PQm

3

2

1.5

1.1

1.01

0

0.5

0.9

0.99

3

1

)1()(limlim)1(

1

,1

1

t

stsvv

t

t

Avgt

where t is an arbitrary time 1t .

Fill the following table in order to determine the instantaneous velocity at 1t for 29.4)( tts :

Note: You may remark that the calculations performed in problems A and B are very similar. This is because we can

interpret the instantaneous velocity )1(v (at 1t ) as the slope of the tangent line to the graph of )(ts at the point

)9.4,1(P .

Homework: from Exercise Set 2.1: problems 3,4,5,6 and 8.

1.2 The limit of a function (intuitive approach):

In the previous section we used the notation PQPQmm

lim and t

Avgt

vv ,1

1lim)1(

and these notions were introduced

intuitively. However, what is the limit of a general function )(lim xfax

, and how do we calculate such a limit?

As suggested by the previous examples, our first intuitive definition of )(lim xfax

is:

Definition 1: )(lim xfax

is the value L (if it exists) which )(xf approaches when ax appoaches , but ax .

This definition offers us a simple but non-rigorous method to calculate limits.

Example 1: Consider 2)( 2 xxxf . Calculate )(lim2

xfx

by building a table of value as below, and infer from

this table what the value of this limit should be:

Then graph )(xf to confirm your claim..

Note that the value of )(lim2

xfx

is in general totally unrelated to )2(f ,

although in many particular cases, these two values coincide.

t Time Interval

tAvgv ,1

2 ]2,1[

1.5 ]5.1,1[

1.1 ]1.1,1[

1.01 ]01.1,1[

1.001 ]001.1,1[

x 2)( 2 xxxf

1

1.5

1.9

1.99

1.999

3

2.5

2.1

2.01

2.001

4

Example 2: Calculate 1

1lim

2

1

x

x

x using a table method as above. Note that in this case

1

1)(

2

x

xxf is undefined at

1. In fact, )(xf can take any value at 1x without changing the value of )(lim1

xfx

.

A slightly more precise definition of )(lim xfax

:

Definition 2: We say that Lxfax

)(lim if (all values of) )(xf are (or can be made) arbitrarily close (or as close as

we desire) to L when x is close enough to a.

What this says is that the entire graph of the function )(xf is situated in an arbitrarily narrow band around L , when

x is in a narrow enough band around a.

Example 3: Using the method shown in Examples 1 and 2, and keeping in mind Definition 2, calculate:

x

x

x

)sin(lim

0

Then graph x

xxf

)sin()( to confirm or clarify your answer.

Example 4: Estimate 2

2

0

39lim

x

x

x

Then graph 2

2 39)(

x

xxf

to confirm or clarify your answer.

5

Example 5: Estimate, if possible:

xx

sinlim

0

Example 6: The function

0for ,1

0for ,0

)(

x

and

x

xf is graphed below:

Figure 3: Graph of the function )(xf given above;

What is )(lim0

xfx

? We see that as 0)( : 0 with 0 xfxx , but as 1)( : 0 with 0 xfxx , and we write:

1)(lim and 0)(lim00

xfxfxx

; hence )(lim0

xfx

does not exist in this case, since there is no single number L such

that )(xf is arbitrarily close to L (or which )(xf approaches) when x is close enough to 0 .

Definition 3 :

a) We write )(lim xfax

=L if we can make the values of )(xf arbitrarily close to L when x is close enough to a, but

ax .

b) We write )(lim xfax

=L if we can make the values of )(xf arbitrarily close to L when x is close enough to a, but

ax .

Theorem 1: Lxfax

)(lim )(lim xfax

= )(lim xfax

=L .

Therefore, if (1) )(lim xfax

)(lim xfax

, then )(lim xfax

does not exist.

Condition (1) is one of the easiest (and frequent) way to show that )(lim xfax

does not exist.

6

Example 7: Do Example 7 on page 92 in the textbook.

Homework: Do problems 1,2,4,6,8,10,11,20,23,25 from Exercise Set 2.2

1.4. The precise definition of limit:

The intuitive definition of limit introduced in 1.2 by definitions 1 and 2 is too vague and inadequate in many cases.

This is because the main terms which are used by this definition: )(xf approaches L (or )(xf is arbitrarily close to

L) when x is close enough to a are too vague and need to be clearly defined and quantified.

For example, how can we be sure that 1)sin(

lim0

x

x

x and not 9999999.0

)sin(lim

0

x

x

x ? and so on for

the other limits.

In order to set limits on a rigorous and precise basis, consider the simple function 12)( xxf , for which we want

to determine )(lim3

xfx

. Intuitively (from a graph of )(xf or from a table) we guess that 5)(lim3

xfx

, but how can

we be sure?

Definition 1: (the precise definition of the limit):

We say that Lxfax

)(lim if for any 0 there exist a 0 such that Lxf )( when ax0 .

Note: We see that this definition expresses rigorously the idea of Definition 2 in 1.2 by quantifying the concept of

“arbitrarily close” with and of “close enough” with . We sometimes even say - close ( as we think of as a

very small number, or a challenge required for the function )(xf ).

For the function 12)( xxf , the illustration of Definition 1 is given in Figure 1 below.

Think of as a variable challenge for )(xf , which determines a

changing corresponding .

Figure 1: Illustration of 512lim3

xx

7

In order to rigorously prove that 512lim3

xx

according to definition 1, we need to express 512x in

terms of 3x . But we see that 2

33262512

xxxx , and therefore

2

in this case.

For confirmation, verify that

5122

3 xx , which proves that 512lim3

xx

in terms of the

definition 1.

So, in general, in order to show rigorously that Lxfax

)(lim we need to find for a given such that

ax0 Lxf )( .

Example 2: Show rigorously that 754lim3

xx

, that is find such that 7x 754x

for any challenge .

Draw a graph of 54)( xxf and confirm your finding.

The illustration of the general case Lxfax

)(lim is shown in Figure 2 below:

Figure 2: Illustration of Lxfax

)(lim

8

Example 3: Show rigorously that 9lim 2

3

x

x

Example 4: Use a graphical method (or otherwise, as above) to show that 265lim 3

1

xx

x

We often need to calculate rigorously and find side limits only.

Definition 2:

a) We say that Lxfax

)(lim if for any 0 there exist a 0 such that Lxf )( when axa (that is

when x is close enough to a but at the left of a ) ;

b) We say that Lxfax

)(lim if for any 0 there exist a 0 such that Lxf )( when axa (that is

when x is close enough to a but at the right of a ) ;

Example 5:

Show that 0lim0

xx

(note that there is no left side limit in this case, and therefore we need to use the right side

limit only)

Homework: Do problems 1,3,4,5,7,13,14,19,21,22, 23,24,26,27,29,32,35 and 36 from Exercise Set 2.4

9

1.3. Limit laws. Limits of elementary functions and the squeeze theorem.

A. Limits Laws. Limits of elementary functions:

Goals:

Show an efficient yet rigorous method to calculate limits: show that afxfax

)(lim if )(xf is an elementary

function and if )(af exists .

Solve limits of the form:

0

0 ;

Lesson:

Theorem 1 (Limit Laws):

Consider 212211 and : and : DDaRDgRDf such that MxgLxf

axax

)(lim and )(lim . Then the following

limits exist and are equal to the indicated values:

(1)

even) isn (when 0 and any for L)(lim

any for )(lim

)()(lim

0for )(

)(lim

)()(lim

)()(lim

)()(lim

any for )(lim

LNnxf

NnLxf

MLxgxf

MM

L

xg

xf

MLxgxf

MLxgxf

MLxgxf

RkLkxfk

nn

ax

nn

ax

ax

ax

ax

ax

ax

ax

Proof:

These laws can be shown using the definition 1 in 1.4, and the methods of section 1.4. Some of these proofs are

harder than others, and are shown in Appendix F of the textbook.

Let us prove 1)1( for illustration purposes.

We have that

(2) for any 01 there exist a 01 such that 1)( Lxf when 10 ax

And we have to show that

(3) for any 02 there exist a 02 such that 2)( Lkxfk when 20 ax .

10

Let us consider 0k (if k=0 then the result and its proof are trivial since it says that 00lim ax

) . Take in (2) 21

k

and in (3) choose 12 . Then we see that (3) is true when (2) is true.

Q.E.D.

From now on, we consider properties (1) true and we use them in problems. Using these, calculate:

Example 1:

a) 32lim 2

5

xx

x b)

1

2lim

2

2

3

x

x

x c)

1

1lim

2

x

x

x

We see that the method to calculate limits ultimately amounts now to substituting )(in with xfax when )(af

exists.

We state this clearly in general:

Theorem 2 (substitution):

a) If xPxf n)( is a polynomial of degree n then aPxP nnax

lim .

b) If xQ

xPxf

m

n)( is a rational function (ratio of two polynomials) then

aQ

aP

xQ

xP

m

n

m

n

ax

lim if 0aQm .

c) If )( xf is an algebraic function (a combination of , polynomials and rational functions by using

addition, subtraction, multiplication or composition), then afxfax

)(lim if )(af exists (or, in other words,

if a is in the domain of f(x)) .

Proof: The proof of a) and b) are direct consequences of Theorem 1. The proof of part c) is given in Appendix F in

the textbook (listed as Theorem 8).

Example 2:

Using Theorem 2, calculate the limits:

a) 23

12lim

2

2

x

x

x b) 635lim 23

3

xxx

x .

Do also Example 1 (Page 99, of Section 2.3) in the textbook.

11

The next theorem will help us solve cases of study of the form

0

0 .

Theorem 3: If )()( xgxf for all Ix , except for Ia , then )(lim)(lim xgxfaxax

.

This theorem allows us to solve limits the form

0

0 , by first algebraically simplifying the function (usually by

cancelling an pax factor from both top and bottom of the fraction) and then calculating the limit of the simplified

function.

Example 3: Using Theorems 3 and 2 above, as well as Theorem 1 in 1.2 calculate the limits:

a) 1

1lim

21

x

x

x b)

h

h

h

9)3(lim

2

0

c)

x

x

x

416lim

2

0

d) If )(lim calculate 4 if 28

4 if 4)(

4xf

xx

xxxf

x

. Then sketch a graph of this function to confirm your result.

e) Calculate xx 0lim

f) Calculate x

x

x 0lim

g) 32

132lim

2

2

1

xx

xx

x h)

1

1lim

3

4

1

t

t

t i)

xxxx 20

11lim

j) 32

9lim

2

2

3

xx

x

x k)

xx

xxx

x

2

23

0lim l)

23

1

2

2lim

22 xxxxx

m) x

xx

x

33lim

0 n)

24

11lim

0

x

x

x

B. The squeeze theorem:

New types of limits (including limits of trigonometric functions and combinations of these) can be calculated using

an inequality result involving limits called the squeeze theorem.

This theorem is based on:

Theorem 4: If )()( xgxf when x is near a (except possibly at a ) and MxgLxfaxax

)(lim and )(lim , then :

(4) ML .

Proof: See the proof of Theorem 2 in Appendix F.

Based on this theorem, we can easily prove the following:

12

Theorem 5 (the squeeze theorem):

If )()()( xhxgxf when x is near a (except possibly at a ) and if )(lim)(lim Lxhxfaxax

, then : )(lim Lxgax

.

Proof: Based on above, if Mxgax

)(lim , then LML , which shows that ML . Q.E.D.

The squeeze theorem can be illustrated graphically and in a diagram as below

)( )( )( xhxgxf

L

Figure 1: Illustration of the squeeze theorem. Figure 2: Diagram of the squeeze theorem

This theorem is especially useful in calculating limits of trigonometric functions.

First, based on it, we can prove:

Theorem 6 (limits of trigonometric functions):

If )(xf is a trigonometric function (one of the functions )cot(or )tan( ),cos( ),sin( xxxx ) or a combination (using

addition, subtraction, multiplication or composition of these functions or a combination of these functions and the

algebraic functions mention in Theorem 2) then

(5) afxfax

)(lim if )(af exists (or, in other words, if a is in the domain of f(x)) .

Proof: see the proofs on page 123 of the textbook.

Note: This theorem, together with Theorem 2 above, allows us to do substitution for all “elementary functions”

(algebraic, trigonometric or combinations of these) as long as a is in the domain of the function.

Example 4: Calculate the following limits:

a) 2

0sinlim x

x b)

x

x

x

22

0sinlim

However, the squeeze theorem is used directly for a different type of problems, where direct evaluation of the

function is impossible:

Example 5: Using the squeeze theorem 5, show that 01

sinlim 2

0

xx

x (see Example 11 on page 105).

Homework: Problems 1, 2, 4,9,13,15,17,19,22,25,27,28,29,32,33,36,40,41,42,48,50 and 63 from Exercise Set 2.3 .

13

1.6. Limits at infinity and infinite limits:

So far we are able to calculate afxfax

)(lim if a is in the domain of )(xf , or for a case of study of the form

0

0.

In this section we will study new important types of limits involving .

A. Infinite limits:

Definition 1:

a)

)(lim xfax

if the values of )(xf can be made arbitrarily large when x is close enough to a , or in symbols, if for

any 0A (no matter how large !) there exist a 0 such that Axf )( when ax0 .

b)

)(lim xfax

if the values of )(xf can be made arbitrarily large negative when x is close enough to a , or in

symbols, if for any 0A (no matter how large !) there exist a 0 such that Axf )( when ax0 .

Example 1:

a) Graph 2

1)( and

1)(

xxg

xxf and infer infinite limit results from the graph.

b) Using the definition above, show that

1

lim and 1

lim , 1

lim ,1

lim00

2020

xxxx xxxx

c) What is

xx

1lim

0 ?

Note: We generalize the results from the example above and state:

(1) 0

1 and

0

1 . These two results mean that

0)(lim if

)(

1lim xf

xf axax

and

0)(lim if

)(

1lim xf

xf axax (of course, in these results a could be replaced by

aa or , as needed).

Also

0)(lim xf

ax means that as x approaches a, f(x) approaches 0 but only with positive values (f(x) is positive as x

approaches a), and

0)(lim xf

ax means that as x approaches a, f(x) approaches 0 but only with negative values (f(x) is

negative as x approaches a) .

The result (1) is the most important tool to calculate infinite limits. Therefore, in order to evaluate a limit of the form

0

1,

we evaluate the function, and determine the sign of the denominator (best by using a sign table for it !!), then use (1).

14

Example 2:

Calculate the limits:

a) 3

2lim

3 x

x

x and

3

2lim

3 x

x

x , therefore ?

3

2lim

3

x

x

x

b) 23 3

2lim

x

x

x

c) )tan(lim

2

xx

and )tan(lim

2

xx

, therefore ?)tan(lim

2

xx

d) Do problems 29, 31 and 35 from Exercise Set 2.2 in the textbook.

In the previous examples, if )(or )(lim

xfax

, or if )(or )(lim

xfax

or if )(or )(lim

xfax

, then

the line ax is called a Vertical Asymptote (V.A.) for the graph of )(xf (see, for example, the limits in Example 1).

Therefore:

Definition 2: The line ax is called a Vertical Asymptote (V.A.) for the graph of )(xf if at least one of the following

six cases takes place:

)(or )(lim

xfax

, or )(or )(lim

xfax

or )(or )(lim

xfax

.

Practically, to calculate the V.A. for a given function, we calculate first the zeros of the denominator (if the function

is a fraction) and then calculate the limit of the function at these points.

Example 3:

a) Problem 38 from problem set 2.2 in the textbook.

b) Find all the vertical asymptotes of :

i) 95

8)(

xxxf ii)

4310

7)(

x

xxf

iii)

x

xxf

)cos()(

iv) x

xxf

cos)( v)

x

xxf

)sin()( vi)

x

xxf

2sin

)tan()(

B. Limits at infinity:

We turn now to the (last main) case when x .

Definition 3:

a) Lxfx

)(lim if for any 0 there exist a 0B such that Lxf )( when Bx

(or in words if )(xf is arbitrarily close to L when x is large enough). Alternately:

b) Lxfx

)(lim if for any 0 there exist a 0B such that Lxf )( when Bx

(or in words if )(xf is arbitrarily close to L when x is large enough and negative).

15

c) Definition 1 can be combined with Definition 3 (use this as Exercise) to produce:

)(or )(lim and )(or )(lim

xfxfxx

.

Example 4:

a) Calculate, using the definition above: xx

1lim

and

xx

1lim

b) Graph these functions to confirm your result .

c) Calculate 2

1lim

xx .

Note: We generalize the results from the examples above and state:

(2) 01

This means that

)(lim if 0

)(

1lim xf

xf axax (which is relatively easy to prove using the

Definitions 1. a from this section and Definition 1 in 1.4.

From now on, we will use the result (2) as a main tool to calculate limits at .

Example 5: Calculate the following limits by using (2) above:

a) 2

lim x

x

x b)

23lim

2

2

xx

x

x c)

32

1lim

xx

d) xxxx

39lim 2

e) xxxx

2lim 2

The following general results involving will facilitate the direct substitution of in limits:

(3)

)(lim if ,

)(lim if ,

if ,

xga

xga

Ra

a

x

x (4)

0)(lim if ,0

0 if ,

0 if ,

xga

a

a

a

x

(5)

0)(lim if ,

0 if ,0

0 if ,

0 xgn

n

n

x

n

(6)

)(lim if ,

0 if ,

0 if ,

0 if ,

xga

a

a

a

a

x

16

These results can be proved rigorously with a similar approach as described in (1) and (2). Understand them by

thinking of as a quantity larger than any real number, memorize especially the formulas (1) to (6) , and solve the

following problems by substitution:

Example 6:

a) 4

53lim

x

x

x b)

532

lim

2

3

xx

xxx

x c)

1

9lim

3

6

x

xx

x

d) bxxaxxx

22lim e) 1lim 2

xx

f) 2

3lim

3

24

xx

xxx

x

g) x

xearctanlim

h) xx

xx

x ee

ee33

33

lim

i) xx

xx

x ee

ee33

33

0lim

j) 1

)(sinlim

2

2

x

x

x k) )cos(lim 2 xe x

x

l) 32lim

3

xx

x

m) 3

2lim

3

x

x

x n)

35 5lim

x

e x

x o)

65

82lim

2

2

2

xx

xx

x

p) )cot(lim xx

Definition 4:

a) The line by is called a Horizontal Asymptote (H.A.) at for the graph of )(xf if

bxfx

)(lim

b) The line by is called a Horizontal Asymptote (H.A.) at for the graph of )(xf if

bxfx

)(lim

Therefore, a V.A. involve a finite x value and an infinite f(x), while a H.A. involve an infinite x value and a finite

f(x) value there.

In problems, to calculate all H.A. for a given function, calculate )(lim and )(lim xfxfxx

. If one of these limits

produce a finite value, then the function )(xf has a H.A. there, otherwise it doesn’t.

Example 7:

Find the Horizontal Asymptotes at at and for the following functions. Then graph these functions to

visualize your result:

17

a) 1

1)(

2

2

x

xxf b)

2

12)(

2

2

xx

xxxf c)

56)(

2

3

xx

xxxf

Homework:

What problems are left from above, and problems 19, 21,22,38,41,44 and 45 from the Exercise Set 2.6 in the

textbook.

1.5 Continuity of functions:

Functions which have the property that )()(lim afxfax

are called continuous at a . Continuity of functions is a

very important property which some functions verify on an entire interval or on their entire domain of definition.

Functions which are continuous on an interval will satisfy other theorems (such as the intermediate value theorem)

which allows us to find their zeros.

In this section we will learn tools (definitions and theorems) to establish the continuity of functions at a point and on

an entire interval, as well as the intermediate value theorem for continuous functions.

Definition 1: A function RDf : is continuous at a number Da if (1) )()(lim afxfax

.

Note that (1) actually says 3 things about the function )(xf . It says that:

a) )(af exists ( )(xf is defined at a ) ;

b) )(lim xfax

exists and

c) )()(lim afxfax

.

Remembering the meaning of )(lim xfax

, we realize that (1) represents the intuitive fact that )(xf does not have a

jump (or a break – discontinuity) at a . Therefore, when we trace the graph of )(xf close to a , the graph passes

continuously through the point )(, afa .

Definition 2:

a) A function which is not continuous at a (that is, which fails to satisfy one of the conditions a), b) or c)

above) is called discontinuous at a .

b) A function for which the condition b) above is satisfied, but one of a) and c) are not satisfied is said to have a

removable discontinuity at a , as in this case the function )(xf can be re-defined at a to make it

continuous there.

Example 1: See Example 1 on page 119 in the textbook.

Example 2: Determine the continuity of the following functions at the indicated point:

a) 2

2)(

2

x

xxxf at 2a b) 0at

0for 1

0 if )(

2

a

x

xxxf

18

c) 0at

0for 0

0 if )(

a

x

xx

x

xf . Then graph these functions to visualize your claim.

Note that some functions (such as xxf )( ) are not defined to the left of 0a , and therefore only a side

continuity can be defined there. Side continuity is also useful to define continuity on an entire interval.

Definition 2:

a) We say that RDf : is continuous from the right at a number Da if (2) )()(lim afxfax

b) ) We say that RDf : is continuous from the left at a number Da if (3) )()(lim afxfax

Example 3: Show that xxf )( is continuous from the right at 0a . Can this function be continuous from the

left at 0a ?

Definition 3:

a) A function Rbaf ,: is continuous on the entire interval ba, is it is continuous at each number

bac , and it is right continuous at a and left continuous at b ;

b) A function Rbaf ,: is continuous on the entire interval ba, is it is continuous at each number

bac , ;

Example 4:

Show that 211)( xxf is continuous on 1,1 .

Following the model of this exercise (and remembering Theorems 2 and 6 in 1.3), we realize that all elementary

functions (functions which are combinations of , polynomials, rational and trigonometric functions) are

continuous at each number in their domain. We state this clearly in the next theorem (which uses similar results for

the other types of functions mentioned):

Theorem 1: The following types of functions are continuous at every number in their corresponding domains:

Polynomials, Rational Functions, Root Functions (Irrational Functions), Trigonometric Functions, Inverse

Trigonometric Functions, Exponential Functions, Logarithmic Functions and combinations of these functions (using

+ , - , multiplication, division or composition ).

Example 5:

a) Where is 1

)ln()(

2

x

xxf continuous and where is it discontinuous?

Hint: Find the domain of )(xf first, then use the result from Theorem 1.

b) Do problems 15, 17 and 22 from Exercise Set 2.5 in the textbook.

19

Theorem 2 (The Intermediate Function Theorem):

Let Rbaf ,: be a continuous function on the entire interval ba, , and let u be an arbitrary number strictly

between )(af and )(bf , with )()( bfaf . Then there exists (at least) a number bac , such that ucf )( .

Note: What Theorem 2 suggests is that continuous functions achieve (with a particular )(cf ) all values in their

range (if their range is an interval) . The proof of this theorem can be found in more advanced Calculus texts, and it

is based on Cauchy’s completion axiom (see a model of the proof on Wikipedia).

We Use Theorem 2 to find zeros of a continuous function, or to prove that such zeros exist.

Example 6:

Show that there is a root of the equation 02364 23 xxx between 1 and 2. Can we reduce the length of the

interval which contains this root? Can you suggest an algorithm to find the root to a desired precision?

Homework:

What left from above and problems 3,4,5,6,7, 16, 18, 19, 20, 21,4042,45,46,51 and 54.

1.7 Review:

In this chapter, we learnt

- Limits at an intuitive and motivational level (Sections 1.1 and 1.2) ;

- Rigorous methods to calculate limits (Section 1.4)

- Practical methods to calculate limits (Sections 1.3 and 1.6)

- Continuity of functions based on limits (Section 1.5) .

Learn the main results (definitions, theorems and formulas from these sections) (best write a review guide to include

these) , then solve the following problems:

- In the textbook, do the Review problems from pages 165 to 168, except the problems involving the

derivative of a function or rates of change (problems 11,12,13,14 from Concept Check, problems 19,20,21

from True and False, and problems 38 to 50 from Exercises).

- Do the problems below:

1. Calculate xx

coslim5.0

by building a table of values xx cos, around 0.5 (the intuitive approach of guessing

the limit using a table (used in 1.2) ).

2. Prove the following limits rigorously (using the rigorous definition of limit, as in 1.4):

a) 23

11lim

3

x

x b) 0lim 2

0

x

x c) 154lim 2

2

xx

x

20

3. Calculate the following limits (using methods from the sections 1.3 and 1.6):

a) 5

56lim

2

5

x

xx

x b)

43

4lim

2

2

1

xx

xx

x c)

h

h

h

8)2(lim

3

0

d) 1

1lim

3

4

1

x

x

x e)

8

2lim

32

x

x

x f)

xxxx 20

11lim

g) x

xx

x

11lim

0 h)

20 16

4lim

xx

x

x

i)

x

x

x

11

0

3)3(lim

j) 1

1lim

3

2

xx

x

x k)

12

23lim

x

x

x l)

542

264lim

3

23

xx

xx

x

m) 1

9lim

3

6

x

xx

x n)

1

9lim

3

6

x

xx

x o) xxx

x39lim 2