chapter 11: sequences; indeterminate forms, improper integrals

Main MenuSalas, Hille, Etgen Calculus: One and Several Variables

Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.

Section 11.1 The Least Upper Bound Axioma. Least Upper Bound Axiomb. Examplesc. Theorem 11.1.2d. Examplee. Greatest Lower Boundf. Theorem 11.1.4

Section 11.2 Sequences of Real Numbersa. Definitionb. Laws of Formationc. Types of Sequencesd. Range of a Sequencee. Increasing, Decreasing, Monotonic Sequencesf. Example

Section 11.3 Limit of A Sequencea. Definitionb. Examplec. Uniqueness of Limit Theorem d. Convergent and Divergent Sequencese. Convergence Theoremf. Exampleg. Theorem 11.3.7h. Pinching Theorem for Sequencesi. Corollaryj. Continuous Functions Applied to Convergent Sequences

Chapter 11: Sequences; Indeterminate Forms; Improper IntegralsSection 11.4 Some Important Limitsa. Common Limitsb. Limit Propertiesc. More Properties

Section 11.5 The Indeterminate Form (0/0)a. L’Hôpitals Rule (0/0)b. Examplec. The Cauchy Mean-Value Theorem

Section 11.6 The Indeterminate Form (∞/∞); Other Indeterminate Forms

a. L’Hôpitals Rule (∞/∞)b. Limit Propertiesc. Exampled. Indeterminates 00, 1∞, ∞0

e. Example

Section 11.7 Improper Integralsa. Integrals Over Unbounded Intervalsb. Examplesc. Limit Propertyd. Comparison Teste. Integrals of Unbounded Functionsf. Example



The Least Upper Bound Axiom




We indicate the least upper bound of a set S by writing lub S. As you will see from the examples below, the least upper bound idea has wide applicability.

(1) lub (−∞, 0) = 0, lub(−∞, 0] = 0.

(2) lub (−4,−1) = −1, lub(−4,−1] = −1.

(3) lub {1/2, 2/3, 3/4, . . . , n/(n + 1), . . . } = 1.

(4) lub {−1/2,−1/8,−1/27, . . . ,−1/n3, . . . } = 0.

(5) lub {x : x2 < 3} = lub{x : } =

(6) For each decimal fraction

b = 0.b1b2b3, . . . ,

we have

b = lub {0.b1, 0.b1b2, 0.b1b2b3, . . . }.

(7) If S consists of the lengths of all polygonal paths inscribed in a semicircle of radius 1, then lub S = π (half the circumference of the unit circle).

3 3x− < < 3




Example

(a) Let S = {1/2, 2/3, 3/4, . . . , n/(n + 1), . . . } and take ε = 0.0001. Since 1 is the least upper bound of S, there must be a number s ∈ S such that

1 − 0.0001 < s < 1.

There is: take, for example,

(b) Let S = {0, 1, 2, 3} and take ε = 0.00001. It is clear that 3 is the least upper bound of S. Therefore, there must be a number s ∈ S such that

3 − 0.00001 < s ≤ 3.There is: s = 3.

99,999 .100,000

s =



Sequences of Real Numbers



Sequences of Real NumbersSuppose we have a sequence of real numbers

a1, a2, a3, . . . , an, . . .

By convention, a1 is called the first term of the sequence, a2 the second term, and so on. More generally, an, the term with index n, is called the nth term.Sequences can be defined by giving the law of formation. For example:

2

1 1 12 3 4

1 2 3 42 3 4 5

1 is the sequence 1, , , , . . .

is the sequence , , , , . . .1

is the sequence 1, 4, 9, 16, . . .

n

n

n

an

nbn

c n

=

=+=

It’s like defining f by giving f (x).




Sequences can be multiplied by constants; they can be added, subtracted, andmultiplied. From

a1, a2, a3, . . . , an, . . . and b1, b2, b3, . . . , bn, . . .

we can form

the scalar product sequence : αa1, αa2, αa3, . . . , αan, . . . ,

the sum sequence : a1 + b1, a2 + b2, a3 + b3, . . . , an + bn, . . . ,

the difference sequence : a1 − b1, a2 − b2, a3 − b3, . . . , an − bn, . . . ,

the product sequence : a1b1, a2b2, a3b3, . . . , anbn, . . . .

If the bi ’s are all different from zero, we can form

the quotient sequence :

the reciprocal sequence :1 2 3

1 1 1 1, , , . . . , , . . .nb b b b

31 2

1 2 3

, , , . . . , , . . .n

n

a aa ab b b b




The range of a sequence is the set of values taken on by the sequence. While there can be repetition in a sequence, there can be no repetition in the statement of its range. The range is a set, and in a set there is no repetition. A number is either in a particular set or it’s not. It can’t be there more than once. The sequences

0, 1, 0, 1, 0, 1, 0, 1, . . . and 0, 0, 1, 1, 0, 0, 1, 1, . . .

both have the same range: the set {0, 1}. The range of the sequence0, 1,−1, 2, 2,−2, 3, 3, 3,−3, 4, 4, 4, 4,−4, . . .

is the set of integers.



Sequences of Real NumbersMany of the sequences we work with have some regularity. They either have an upward tendency or they have a downward tendency. The following terminology is standard. The sequence with terms an is said to be

increasing if an < an+1 for all n,

nondecreasing if an ≤ an+1 for all n,

decreasing if an > an+1 for all n,

nonincreasing if an ≥ an+1 for all n.

A sequence that satisfies any of these conditions is called monotonic.The sequences

1, ½, 1∕3, ¼, . . . , 1∕n , . . .

2, 4, 8, 16, . . . , 2n, . . .

2, 2, 4, 4, 6, 6, . . . , 2n, 2n, . . .

are monotonic. The sequence

1, ½, 1, 1∕3, 1, ¼, 1, . . .

is not monotonic.



Sequences of Real NumbersExampleThe sequence

is increasing. It is bounded below by ½ (the greatest lower bound) and above by 1 (the least upper bound).ProofSince

we have an < an+1. This confirms that the sequence is increasing. The sequence can be displayed as

It is clear that ½ is the greatest lower bound and 1 is the least upper bound.

1nna

n=

+

( ) ( )( )

21

2

1 / 2 1 1 2 1 1/ 1 2 2

n

n

n na n n n na n n n n n n+ + + + + + += = ⋅ = >

+ + +

1 2 3 4 98 99, , , ,. . . , , , . . .2 3 4 5 99 100



Limit of A Sequence



Limit of A SequenceExampleSince

it is intuitively clear that

To verify that this statement conforms to the definition of limit of a sequence, we must show that for each ε > 0 there exists a positive integer K such thatif n ≥ K, then

To do this, we fix ε > 0 and note that

We now choose K sufficiently large that . If n ≥ K, thenand consequently

2 21 1 1

nn n

=+ +

2lim 21n

nn→∞

=+

2 21

nn

ε− <+

( )2 2 121 1

2 2 221 1

n nnn n n n n

− +

+ +

−− = = = <

+ +

2 221

nn n

ε− < <+

2 K ε< 2 2n K ε≤ <



Limit of A Sequence



Limit of A Sequence

ExampleWe shall show that the sequence is convergent. Since

3 = (3n)1/n < (3n + 4n)1/n < (4n + 4n)1/n = (2 · 4n)1/n = 21/n · 4 ≤ 8,

the sequence is bounded. Note that

(3n + 4n)(n+1)/n = (3n + 4n)1/n(3n + 4n)= (3n + 4n)1/n3n + (3n + 4n)1/n4n.

Since(3n + 4n)1/n > (3n)1/n = 3 and (3n + 4n)1/n > (4n)1/n = 4,

we have(3n + 4n)(n+1)/n > 3 · (3n) + 4 · (4n) = 3n+1 + 4n+1.

Taking the (n + 1)st root of the left and right sides of this inequality, we obtain

(3n + 4n)1/n > (3n+1 + 4n+1)1/(n+1)

The sequence is decreasing. Being also bounded, it must be convergent.

( )1/3 4

nn nna = +



Limit of A Sequence



Limit of A Sequence

The following is an obvious corollary to the pinching theorem.

A limit to remember



Limit of A Sequence

Continuous Functions Applied to Convergent Sequences



Some Important Limits



The Indeterminate Form (0/0)




ExampleFind

SolutionAs x → 0+, both numerator and denominator tend to 0 and

It follows from L’Hôpital’s rule that

For short, we can write

0lim

sinx

xx+→

( )( ) ( )( )

1 2 0 01coscos 1/ 2

f x xg x xx x

′= = → =

′

0lim 0

sinx

xx+→→

0 0

2lim lim 0sin cosx x

x xx x+ +→ →

=



Other Indeterminate Forms




ExampleDetermine the behavior of as n→∞.

SolutionTo use the methods of calculus, we investigate

Since both numerator and denominator tend to ∞ with x, we try L’Hôpital’s rule:

Therefore the sequence must also diverge to ∞.

2

2n

nan

=

2

2limx

x x→∞

( )2

2

2 ln 22 2 ln 2lim lim lim2 2

xx x

x x xx x→∞ →∞ →∞= ∞




The Indeterminates 00, 1∞, ∞0

Such indeterminates are usually handled by first applying the logarithm function:

y = [ f (x)]g(x) gives ln y = g(x) ln f (x).



Other Indeterminate FormsExample (1∞) Find

SolutionHere we are dealing with an indeterminate of the form 1∞: as x → 0+, 1 + x → 1 and 1/x →∞. Taking the logarithm and then applying L’Hôpital’s rule, we have

As x → 0+, ln (1 + x)1/x → 1 and (1 + x)1/x = eln(1+x)1/x → e1 = e. Set x = 1/nand we have the familiar result: as n→∞, [1 + (1/n)]n → e.

( )1/

0lim 1 x

xx

+→+

( ) ( )1/

0 0 0

ln 1 1lim ln 1 lim lim 11

x

x x x

xx

x x+ + +→ → →

++ = =

+



Improper IntegralsIntegrals over Unbounded IntervalsWe begin with a function f which is continuous on an unbounded interval [a,∞). For each number b > a we can form the definite integral

If, as b tends to ∞, this integral tends to a finite limit L,

( )b

af x dx∫

( )limb

abf x dx L

→∞=∫

then we write

and say that

the improper integral

Otherwise, we say that


converges to L.

diverges.

( )a

f x dx L∞

=∫

( )a

f x dx∞

∫

( )a

f x dx∞

∫



Improper Integrals

As a tends to −∞, sin πa oscillates between −1 and 1. Therefore the integral oscillates between 1/π and −1/π and does not converge.



Improper Integrals



Improper IntegralsIntegrals of Unbounded FunctionsImproper integrals can arise on bounded intervals. Suppose that f is continuous on the half-open interval [a, b) but is unbounded there. For each number c < b, we can form the definite integral

If, as c → b−, the integral tends to a finite limit L, namely, if

( )c

af x dx∫

( )limc

ac bf x dx L

−→=∫

then we write

and say that


Otherwise, we say that the improper integral diverges.

converges to L.

( )b

af x dx L=∫

( )b

af x dx∫



Improper IntegralsExampleTest(∗)

for convergence.

SolutionThe integrand has an infinite discontinuity at x = 2.For integral (∗) to converge both

must converge. Neither does. For instance, as c → 2−,

This tells us that

If we overlook the infinite discontinuity at x = 2, we can be led to the incorrect conclusion that

( )4

21 2dx

x −∫

( ) ( )2 4

2 21 2and

2 2dx dx

x x− −∫ ∫

( )211

1 1 12 22

cc dx

x cx = − = − − →∞ − − −∫

( )2

21 2dx

x −∫ diverges and shows that (∗) diverges.

( )

44

211

1 32 22

dxxx

= − = − − −∫

chapter 11: sequences; indeterminate forms, improper integrals

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