Limits using L’Hopital’s Rule (Sect. 7.5) 0 L’Hˆopital’s ...

[Pages:9]Limits using L'H^opital's Rule (Sect. 7.5)

0 Review: L'H^opital's rule for indeterminate limits .

0 Indeterminate limit . Indeterminate limits ? 0 and - . Overview of improper integrals (Sect. 8.7).

0 L'H^opital's rule for indeterminate limits

0

Remarks:

f (x)

L'H^opital's rule applies on limits of the form L = lim

in

xa g (x )

the case that both f (a) = 0 and g (a) = 0.

0 These limits are called indeterminate and denoted as .

0

Theorem

If functions f , g : I R are differentiable in an open interval containing x = a, with f (a) = g (a) = 0 and g (x) = 0 for x I - {a}, then holds

f (x)

f (x)

lim

= lim

,

xa g (x ) xa g (x )

assuming the limit on the right-hand side exists.

0 L'H^opital's rule for indeterminate limits

0

Example

1 + x - 1 - x/2

Evaluate L = lim

x 0

x2

.

0 Solution: The limit is indeterminate, . But,

0

(1/2)(1 + x)-1/2 - (1/2)

L = lim

.

x 0

2x

0 The limit on the right-hand side is still indeterminate, .

0

We use L'H^opital's rule for a second time,

(-1/4)(1 + x)-3/2 (-1/4)

L = lim

=

.

x 0

2

2

1 We conclude that L = - .

8

0 L'H^opital's rule for indeterminate limits

0

Remark: L'H^opital's rule applies to indeterminate limits only.

Example

1 - cos(x)

Evaluate L = lim

x 0

x + x2

.

0 Solution: The limit is indeterminate . L'H^opital's rule implies,

0

1 - cos(x)

sin(x) 0

L = lim

x 0

x + x2

= lim

=

x0 1 + 2x 1

L = 0.

Remark:

0

0

The limit is not indeterminate, since = 0.

1

1

Therefore, L'H^opital's rule does not hold in this case:

sin(x )

sin(x )

cos(x) 1

lim

= lim

= lim

=.

x0 1 + 2x x0 (1 + 2x ) x0 2

2

Limits using L'H^opital's Rule (Sect. 7.5)

0 Review: L'H^opital's rule for indeterminate limits .

0 Indeterminate limit . Indeterminate limits ? 0 and - . Overview of improper integrals (Sect. 8.7).

Indeterminate limit

Remark: L'H^opital's rule can be generalized to limits

,

and also to side limits.

Example

2 + tan(x)

Evaluate L = lim

.

x

(

2

)-

3

+

sec(x )

Solution: This is an indeterminate limit . L'H^opital's rule implies

2 + tan(x)

sec2(x )

sec(x )

lim

= lim

= lim

x

(

2

)-

3 + sec(x)

x

(

2

)-

sec(x )

tan(x )

x

(

2

)-

tan(x )

sec(x )

1 cos(x)

1

Since

=

=

, then L = 1.

tan(x) cos(x) sin(x) sin(x)

 Indeterminate limit

Remark: Sometimes L'H^opital's rule is not useful.

Example

sec(x )

Evaluate L = lim

.

x

(

2

)-

tan(x )

Solution: We know that this limit can be computed simplifying:

sec(x) = 1 cos(x) = 1

L = 1.

tan(x) cos(x) sin(x) sin(x)

We now try to compute this limit using L'H^opital's rule.

Indeterminate limit

Example

sec(x )

Evaluate L = lim

.

x

(

2

)-

tan(x )

Solution: This is an indeterminate limit . L'H^opital's rule implies

(sec(x ))

sec(x) tan(x)

tan(x )

L = lim

= lim

x

(

2

)-

(tan(x ))

x

(

2

)-

sec2(x )

= lim

.

x

(

2

)-

sec(x )

The later limit is once again indeterminate, . Then

(tan(x ))

sec2(x )

sec(x )

L = lim

= lim

= lim

.

x

(

2

)-

(sec(x ))

x

(

2

)-

sec(x )

tan(x )

x

(

2

)-

tan(x )

L'H^opital's rule gives us a cycling expression.

 Indeterminate limit

Example

3x2 - 5

Evaluate

L

=

lim

x

2x 2

-

x

+

. 3

Solution: This is an indeterminate limit . L'H^opital's rule implies

(3x2 - 5)

6x

6

L

=

lim

x

(2x 2

-

x

+

3)

= lim

= lim

x 4x - 1 x

4

-

1 x

.

1

6

Recalling lim = 0, we get that L = . We conclude that

x x

4

3x2 - 5

3

lim

x

2x 2

-

x

+

3

=

. 2

Limits using L'H^opital's Rule (Sect. 7.5)

0 Review: L'H^opital's rule for indeterminate limits .

0 Indeterminate limit . Indeterminate limits ? 0 and - .

Overview of improper integrals (Sect. 8.7).

Indeterminate limits ? 0 and - .

Remark: Sometimes limits of the form ? 0 and ( - ) can be

0 converted by algebraic identities into indeterminate limits or

0

Example

11

Evaluate L = lim

-.

x0 sin(x) x

Solution: This is a limit of the form ( - ). Since

1 1 x - sin(x)

0

-=

indeterminate .

sin(x) x x sin(x)

0

Then L'H^opital's rule in this case implies

x - sin(x)

1 - cos(x)

L = lim

= lim

x0 x sin(x)

x0 sin(x) + x cos(x)

Indeterminate limits ? 0 and - .

Example

11

Evaluate L = lim

-.

x0 sin(x) x

1 - cos(x)

Solution: Recall L = lim

.

x0 sin(x) + x cos(x)

0 This limit is still indeterminate . Hence

0

1 - cos(x)

sin(x )

0

L = lim

= lim

= = 0.

x0 sin(x) + x cos(x) x0 2 cos(x) - x sin(x) 2

We conclude that L = 0.

Indeterminate limits ? 0 and - .

Example

Evaluate L = lim (3x)2/x .

x

Solution: This limits is of the form 0. So, before using L'H^opital's rule we need to rewrite the function above.

(3x )2/x = eln (3x)2/x

=e

2 x

ln(3x )

.

Since exp is a continuous function, holds

lim (3x )2/x = elimx

2 x

ln(3x )

= elimx

2 ln(3x) x

.

x

The

exponent,

is

an

indeterminate

limit

.

L'H^opital's

rule

implies

2 ln(3x)

2 ln(3x)

2/x

lim

= lim

= lim = 0.

x x

x (x )

x 1

We conclude that L = e0, that is, L = 1.

Limits using L'H^opital's Rule (Sect. 7.5)

0 Review: L'H^opital's rule for indeterminate limits .

0 Indeterminate limit . Indeterminate limits ? 0 and - .

Overview of improper integrals (Sect. 8.7).

Overview of improper integrals (Sect. 8.7)

Remarks:

L'H^opital's rule is useful to compute improper integrals. Improper integrals are the limit of definite integrals when one endpoint if integration approaches ?.

Definition

The improper integral of a continuous function f : [a, ) R is

b

f (x) dx = lim f (x) dx.

a

b a

The improper integral of a continuous function f : (-, b] R is

b

b

f (x) dx = lim f (x) dx.

-

a- a

The improper integral of a continuous function f : (-, ) R,

c

f (x) dx = f (x) dx + f (x) dx.

-

-

c

Overview of improper integrals (Sect. 8.7)

Example

ln(x)

Evaluate I =

1

x2 dx.

Solution: This is an improper integral:

ln(x)

b ln(x)

1

x2

dx = lim

b

1

x2

Integrating by parts, u = ln(x), and dv = dx/x2,

b ln(x)

1

b

x1

1

1

x2

dx =

- x

ln(x) -

1

1

x

- dx x

b ln(x)

ln(b)

b dx

ln(b) 1 b

1

x2 dx = - b

+ 1 x2 = - b

-. x1

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