Chapter 11: Sequences; Indeterminate Forms; Improper Integrals

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

The Least Upper Bound Axiom

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

The Least Upper Bound Axiom

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 =

The Least Upper Bound Axiom

The Least Upper Bound Axiom

Sequences of Real Numbers

Sequences of Real Numbers

Suppose we have a sequence of real numbers

a₁, a₂, a₃, . . . , a_n, . . .

By convention, a₁ is called the first term of the sequence, a₂ the second term, and so on. More generally, a_n, 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).

Sequences of Real Numbers

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

a₁, a₂, a₃, . . . , a_n, . . . and b₁, b₂, b₃, . . . , b_n, . . . we can form

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

the sum sequence : a₁ + b₁, a₂ + b₂, a₃ + b₃, . . . , a_n + b_n, . . . , the difference sequence : a₁ − b₁, a₂ − b₂, a₃ − b₃, . . . , a_n − b_n, . . . ,

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

If the b_i ’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

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.

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 a_n is said to be increasing if a_n < a_n+1 for all n,

nondecreasing if a_n ≤ a_n+1 for all n, decreasing if a_n > a_n+1 for all n, nonincreasing if a_n ≥ a_n+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 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 a_n < a_n+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

Limit of A Sequence

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

2 2 2

2

1 1

n n

n

n n

n n n

− +

+ +

− = = − = <

+ +

2 2

1 2 n

n − < n < ε +

2 K < ε 2 n ≤ 2 K <ε

Limit of A Sequence

Limit of A Sequence

Limit of A Sequence

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 = 2^1/n · 4 ≤ 8, the sequence is bounded. Note that

(3ⁿ + 4^n)(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.

(

)

an = +

Limit of A Sequence

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

Some Important Limits

Some Important Limits

The Indeterminate Form (0/0)

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

→

( ) ( ) (

) (

)

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

+ +

→  → =

The Indeterminate Form (0/0)

Other Indeterminate Forms

Other Indeterminate Forms

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 ln 2

2 2 ln 2

lim lim lim

2 2

x x x

x^→∞ x  x^→∞ x  x^→∞ = ∞

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).

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

( )

0

lim 1

x

+

x

→

+

( )

( )

0 0 0

ln 1 1

lim ln 1 lim lim 1

1

x

x x x

x x

x x

+ + +

→ → →

+ = + =

 +

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

( )

∫

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 Integrals

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

∫

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

x− x−

∫ ∫

( )

1 1

1 1

2 2 1

2

c dx c

x c

x

 

= − −  = − − − → ∞

∫

( )

2 1 2

2 dx x −

∫

( )

4 4

1 2

1

1 3

2 2

2 dx x x

 

= − −  = −

∫