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# Chapter 11: Sequences; Indeterminate Forms; Improper Integrals

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Section 11.1 The Least Upper Bound Axiom

a. Least Upper Bound Axiom

b. Examples

c. Theorem 11.1.2

d. Example

e. Greatest Lower Bound

f. Theorem 11.1.4

Section 11.2 Sequences of Real Numbers

a. Definition

b. Laws of Formation

c. Types of Sequences

d. Range of a Sequence

e. Increasing, Decreasing, Monotonic Sequences

f. Example

Section 11.3 Limit of A Sequence

a. Definition

b. Example

c. Uniqueness of Limit Theorem

d. Convergent and Divergent Sequences

e. Convergence Theorem

f. Example

g. Theorem 11.3.7

h. Pinching Theorem for Sequences

i. Corollary

j. Continuous Functions Applied to Convergent Sequences

### Chapter 11: Sequences; Indeterminate Forms; Improper Integrals

Section 11.4 Some Important Limits

a. Common Limits

b. Limit Properties

c. More Properties

Section 11.5 The Indeterminate Form (0/0)

a. L’Hôpitals Rule (0/0)

b. Example

c. The Cauchy Mean-Value Theorem

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

a. L’Hôpitals Rule (∞/∞)

b. Limit Properties

c. Example

d. Indeterminates 00, 1, ∞0

e. Example

Section 11.7 Improper Integrals

a. Integrals Over Unbounded Intervals

b. Examples

c. Limit Property

d. Comparison Test

e. Integrals of Unbounded Functions

f. Example

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### 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 x 3

− < < 3

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### The Least Upper Bound Axiom

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 =

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### Sequences of Real Numbers

Suppose 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 1 2 3 4 1 2 3 4 2 3 4 5

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

is the sequence , , , , . . . 1

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

n

n

n

a n

b n

n c n

=

= +

=

It’s like defining f by giving f (x).

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### Sequences of Real Numbers

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

, , , . . . , , . . . b b b bn

3 1 2

1 2 3

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

n

a a

a a

b b b b

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### Sequences of Real Numbers

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 sequence

0, 1,−1, 2, 2,−2, 3, 3, 3,−3, 4, 4, 4, 4,−4, . . . is the set of integers.

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### Sequences of Real Numbers

Many 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, 13, ¼, . . . , 1n , . . . 2, 4, 8, 16, . . . , 2n, . . . 2, 2, 4, 4, 6, 6, . . . , 2n, 2n, . . . are monotonic. The sequence

1, ½, 1, 13, 1, ¼, 1, . . . is not monotonic.

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### Sequences of Real Numbers

Example The sequence

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

Proof Since

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.

n 1 a n

= n +

### ( )

2 1

2

1 / 2 1 1 2 1

/ 1 2 2 1

n n

n n

a n n n n

a n n n n n n

+ = + + = + + = + + >

+ + +

1 2 3 4 98 99

, , , ,. . . , , , . . .

2 3 4 5 99 100

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### Limit of A Sequence

Example Since

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 that

if n ≥ K, then

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

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

2 2

1 1 1 n

n = n

+ +

lim 2 2

1

n

n

→∞ n =

+

2 2

1 n

n − <ε +

2 2 1

2

1 1

n n

n

n n

+

+ +

### + +

2 2

1 2 n

n − < n < ε +

2 K < ε 2 n ≤ 2 K

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### Limit of A Sequence

Example

We 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.

3n 4n

1/n

an = +

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### Limit of A Sequence

The following is an obvious corollary to the pinching theorem.

A limit to remember

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### Limit of A Sequence

Continuous Functions Applied to Convergent Sequences

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### The Indeterminate Form (0/0)

Example Find

Solution

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

It follows from L’Hôpital’s rule that

For short, we can write

0

lim sin

x

x

+ x

cos

11/ 2

## )

cos2 10 0

f x x

g x x x x

′ = = → =

′  

0

lim 0

sin

x

x

+ x

0 0

lim lim 2 0

sin cos

x x

x x

x x

+ +

=

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### Other Indeterminate Forms

Example

Determine the behavior of as n→∞.

Solution

To 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 an

= n

2

lim 2

x

x→∞ x

### ( )

2

2

2 ln 2

2 2 ln 2

lim lim lim

2 2

x x x

x→∞ xx→∞ xx→∞ = ∞

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### Other Indeterminate Forms

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).

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### Other Indeterminate Forms

Example (1) Find

Solution

Here 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/n and we have the familiar result: as n→∞, [1 + (1/n)]n → e.

1/

0

x

x

+

1/

0 0 0

x

x x x

+ + +

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### Improper Integrals

Integrals over Unbounded Intervals

We 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

a f x dx

lim b

b a f x dx L

→∞

### ∫

=

then we write

and say that

the improper integral

Otherwise, we say that

the improper integral

converges to L.

diverges.

a f x dx

= L

a f x dx

a f x dx

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### 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.

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### Improper Integrals

Integrals of Unbounded Functions

Improper 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

a f x dx

lim c

c b a

f x dx L

### ∫

=

then we write

and say that

the improper integral

Otherwise, we say that the improper integral diverges.

converges to L.

b

a f x dx = L

b

a f x dx

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### Improper Integrals

Example Test

(∗)

for convergence.

Solution

The 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 1 2

2 dx x

2 4

2 2

1 and 2

2 2

dx dx

xx

### ( )

2

1 1

1 1

2 2 1

2

c dx c

x c

x

 

= − −  = − − − → ∞

2 1 2

2 dx x

### ∫

diverges and shows that (∗) diverges.

4 4

1 2

1

1 3

2 2

2 dx x x

 

= − −  = −

### ∫

In this section we define an integral that is similar to a single integral except that instead of integrating over an interval [a, b], we integrate over a curve C.. Such integrals

But we know that this improper integral is divergent. In other words, the area under the curve is infinite. So the sum of the series must be infinite, that is, the series is..

Thus the given improper integral is convergent and, since the integrand is positive, we can interpret the value of the integral as the area of the shaded region in Figure

Theorem (Comparison Theorem For Functions) Suppose that a ∈ R, that I is an open interval that contains a, and that f,g are real functions defined everywhere on I except possibly at

Suppose that E is bounded, open interval and that each f k.. is differentiable

Then g is defined on [a, b], satifies (11), and is continuous on [a, b] by the Sequential Characterization of Limits.. Thus, f

• Suppose the input graph contains at least one tour of the cities with a total distance at most B. – Then there is a computation path for

Write the following problem on the board: “What is the area of the largest rectangle that can be inscribed in a circle of radius 4?” Have one half of the class try to solve this