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In this section, we will learn:

How to evaluate functions whose values cannot be found at certain points using derivatives.

How to evaluate functions whose values cannot be found at certain points using derivatives.

1.
4

APPLICATIONS OF DIFFERENTIATION

APPLICATIONS OF DIFFERENTIATION

2.
APPLICATIONS OF DIFFERENTIATION

4.4

Indeterminate Forms

and L’Hospital’s Rule

In this section, we will learn:

How to evaluate functions whose

values cannot be found at certain points.

4.4

Indeterminate Forms

and L’Hospital’s Rule

In this section, we will learn:

How to evaluate functions whose

values cannot be found at certain points.

3.
INDETERMINATE FORMS

Suppose we are trying to analyze

the behavior of the function ln x

F ( x)

x 1

Although F is not defined when x = 1,

we need to know how F behaves near 1.

Suppose we are trying to analyze

the behavior of the function ln x

F ( x)

x 1

Although F is not defined when x = 1,

we need to know how F behaves near 1.

4.
INDETERMINATE FORMS Expression 1

In particular, we would like to know

the value of the limit

ln x

lim

x 1 x 1

In particular, we would like to know

the value of the limit

ln x

lim

x 1 x 1

5.
INDETERMINATE FORMS

In computing this limit, we can’t apply

Law 5 of limits (Section 2.3) because

the limit of the denominator is 0.

In fact, although the limit in Expression 1 exists,

its value is not obvious because both numerator

0

and denominator approach 0 and is not defined.

0

In computing this limit, we can’t apply

Law 5 of limits (Section 2.3) because

the limit of the denominator is 0.

In fact, although the limit in Expression 1 exists,

its value is not obvious because both numerator

0

and denominator approach 0 and is not defined.

0

6.
INDETERMINATE FORM —TYPE 0/0

In general, if we have a limit of the form f ( x)

lim

x a g ( x)

where both f(x) → 0 and g(x) → 0 as x → a,

then this limit may or may not exist.

0

It is called an indeterminate form of type .

0

We met some limits of this type in Chapter 2.

In general, if we have a limit of the form f ( x)

lim

x a g ( x)

where both f(x) → 0 and g(x) → 0 as x → a,

then this limit may or may not exist.

0

It is called an indeterminate form of type .

0

We met some limits of this type in Chapter 2.

7.
INDETERMINATE FORMS

For rational functions, we can cancel

common factors:

2

x x x( x 1)

lim 2 lim

x 1 x 1 x 1 ( x 1)( x 1)

x 1

lim

x 1 x 1 2

For rational functions, we can cancel

common factors:

2

x x x( x 1)

lim 2 lim

x 1 x 1 x 1 ( x 1)( x 1)

x 1

lim

x 1 x 1 2

8.
INDETERMINATE FORMS

We used a geometric argument

to show that:

sin x

lim 1

x 0 x

We used a geometric argument

to show that:

sin x

lim 1

x 0 x

9.
INDETERMINATE FORMS

However, these methods do not work

for limits such as Expression 1.

Hence, in this section, we introduce

a systematic method, known as l’Hospital’s Rule,

for the evaluation of indeterminate forms.

However, these methods do not work

for limits such as Expression 1.

Hence, in this section, we introduce

a systematic method, known as l’Hospital’s Rule,

for the evaluation of indeterminate forms.

10.
INDETERMINATE FORMS Expression 2

Another situation in which a limit is

not obvious occurs when we look for

a horizontal asymptote of F and need

to evaluate the limit ln x

lim

x x 1

Another situation in which a limit is

not obvious occurs when we look for

a horizontal asymptote of F and need

to evaluate the limit ln x

lim

x x 1

11.
INDETERMINATE FORMS

It isn’t obvious how to evaluate this limit

because both numerator and denominator

become large as x → ∞.

There is a struggle between the two.

If the numerator wins, the limit will be ∞.

If the denominator wins, the answer will be 0.

Alternatively, there may be some compromise—

the answer may be some finite positive number.

It isn’t obvious how to evaluate this limit

because both numerator and denominator

become large as x → ∞.

There is a struggle between the two.

If the numerator wins, the limit will be ∞.

If the denominator wins, the answer will be 0.

Alternatively, there may be some compromise—

the answer may be some finite positive number.

12.
INDETERMINATE FORM —TYPE ∞/∞

In general, if we have a limit of the form

f ( x)

lim

x a g ( x)

where both f(x) → ∞ (or -∞) and g(x) → ∞

(or -∞), then the limit may or may not exist.

It is called an indeterminate form of type ∞/∞.

In general, if we have a limit of the form

f ( x)

lim

x a g ( x)

where both f(x) → ∞ (or -∞) and g(x) → ∞

(or -∞), then the limit may or may not exist.

It is called an indeterminate form of type ∞/∞.

13.
INDETERMINATE FORMS

We saw in Section 2.6 that this type of limit

can be evaluated for certain functions—

including rational functions—by dividing the

numerator and denominator by the highest

power of x that occurs in the denominator.

For instance, 1

2 1 2

x 1 x 1 0 1

lim 2 lim

x 2 x 1 x 1 20 2

2 2

x

We saw in Section 2.6 that this type of limit

can be evaluated for certain functions—

including rational functions—by dividing the

numerator and denominator by the highest

power of x that occurs in the denominator.

For instance, 1

2 1 2

x 1 x 1 0 1

lim 2 lim

x 2 x 1 x 1 20 2

2 2

x

14.
INDETERMINATE FORMS

This method, though, does not work

for limits such as Expression 2.

However, L’Hospital’s Rule also applies to

this type of indeterminate form.

This method, though, does not work

for limits such as Expression 2.

However, L’Hospital’s Rule also applies to

this type of indeterminate form.

15.
L’HOSPITAL’S RULE

Suppose f and g are differentiable and

g’(x) ≠ 0 on an open interval I that contains a

(except possibly at a).

Suppose lim

x a

f ( x) 0 and lim g ( x) 0

x a

or that lim f ( x) and lim g ( x)

x a x a

In other words, we have an indeterminate form

0

of type or ∞/∞.

0

Suppose f and g are differentiable and

g’(x) ≠ 0 on an open interval I that contains a

(except possibly at a).

Suppose lim

x a

f ( x) 0 and lim g ( x) 0

x a

or that lim f ( x) and lim g ( x)

x a x a

In other words, we have an indeterminate form

0

of type or ∞/∞.

0

16.
L’HOSPITAL’S RULE

f ( x) f '( x)

lim lim

x a g ( x) x a g '( x )

if the limit on the right exists

(or is ∞ or - ∞).

f ( x) f '( x)

lim lim

x a g ( x) x a g '( x )

if the limit on the right exists

(or is ∞ or - ∞).

17.
NOTE 1

L’Hospital’s Rule says that the limit of

a quotient of functions is equal to the limit of

the quotient of their derivatives—provided that

the given conditions are satisfied.

It is especially important to verify the conditions

regarding the limits of f and g before using the rule.

L’Hospital’s Rule says that the limit of

a quotient of functions is equal to the limit of

the quotient of their derivatives—provided that

the given conditions are satisfied.

It is especially important to verify the conditions

regarding the limits of f and g before using the rule.

18.
NOTE 2

The rule is also valid for one-sided limits

and for limits at infinity or negative infinity.

That is, ‘x → a’ can be replaced by any of

the symbols x → a+, x → a-, x → ∞, or x → - ∞.

The rule is also valid for one-sided limits

and for limits at infinity or negative infinity.

That is, ‘x → a’ can be replaced by any of

the symbols x → a+, x → a-, x → ∞, or x → - ∞.

19.
NOTE 3

For the special case in which

f(a) = g(a) = 0, f’ and g’ are continuous,

and g’(a) ≠ 0, it is easy to see why

the rule is true.

For the special case in which

f(a) = g(a) = 0, f’ and g’ are continuous,

and g’(a) ≠ 0, it is easy to see why

the rule is true.

20.
NOTE 3

In fact, using the alternative form of

the definition of a derivative, we have:

f ( x) f (a) f ( x) f (a)

lim

f '( x) f '(a ) x a x a x a

lim lim

x a g '( x ) g '(a ) g ( x) g (a ) x a g ( x) g (a )

lim

x a x a x a

f ( x) f (a)

lim

x a g ( x) g (a)

f ( x)

lim

x a g ( x)

In fact, using the alternative form of

the definition of a derivative, we have:

f ( x) f (a) f ( x) f (a)

lim

f '( x) f '(a ) x a x a x a

lim lim

x a g '( x ) g '(a ) g ( x) g (a ) x a g ( x) g (a )

lim

x a x a x a

f ( x) f (a)

lim

x a g ( x) g (a)

f ( x)

lim

x a g ( x)

21.
NOTE 3

It is more difficult to prove

the general version of l’Hospital’s

It is more difficult to prove

the general version of l’Hospital’s

22.
L’HOSPITAL’S RULE Example 1

ln x

Find lim

x 1 x 1

lim ln x ln1 0 and lim( x 1) 0

x 1 x 1

Thus, we can apply l’Hospital’s Rule:

d

(ln x)

ln x dx 1/ x 1

lim lim lim lim 1

x 1 x 1 x 1 d x 1 1 x 1 x

( x 1)

dx

ln x

Find lim

x 1 x 1

lim ln x ln1 0 and lim( x 1) 0

x 1 x 1

Thus, we can apply l’Hospital’s Rule:

d

(ln x)

ln x dx 1/ x 1

lim lim lim lim 1

x 1 x 1 x 1 d x 1 1 x 1 x

( x 1)

dx

23.
L’HOSPITAL’S RULE Example 2

x

e

Calculate lim 2

x x

We have lim e x and lim x 2

x x

So, l’Hospital’s Rule gives:

d x

x (e ) x

e e

lim 2 lim dx lim

x x x d 2 x 2 x

(x )

dx

x

e

Calculate lim 2

x x

We have lim e x and lim x 2

x x

So, l’Hospital’s Rule gives:

d x

x (e ) x

e e

lim 2 lim dx lim

x x x d 2 x 2 x

(x )

dx

24.
L’HOSPITAL’S RULE Example 2

As ex → ∞ and 2x → ∞ as x → ∞, the limit

on the right side is also indeterminate.

However, a second application of l’Hospital’s

Rule gives: ex ex ex

lim 2 lim lim

x x x 2 x x 2

As ex → ∞ and 2x → ∞ as x → ∞, the limit

on the right side is also indeterminate.

However, a second application of l’Hospital’s

Rule gives: ex ex ex

lim 2 lim lim

x x x 2 x x 2

25.
L’HOSPITAL’S RULE Example 3

ln x

Calculate lim 3

x x

As ln x → ∞ and 3 x as x → ∞,

l’Hospital’s Rule applies: ln x 1/ x

lim lim 1 2 / 3

x 3 x x 3 x

Notice that the limit on the right side

0

is now indeterminate of type .

0

ln x

Calculate lim 3

x x

As ln x → ∞ and 3 x as x → ∞,

l’Hospital’s Rule applies: ln x 1/ x

lim lim 1 2 / 3

x 3 x x 3 x

Notice that the limit on the right side

0

is now indeterminate of type .

0

26.
L’HOSPITAL’S RULE Example 3

However, instead of applying the rule

a second time as we did in Example 2,

we simplify the expression and see that

a second application is unnecessary:

ln x 1/ x 3

lim lim 1 2 / 3 lim 3 0

x 3

x x 3 x x

x

However, instead of applying the rule

a second time as we did in Example 2,

we simplify the expression and see that

a second application is unnecessary:

ln x 1/ x 3

lim lim 1 2 / 3 lim 3 0

x 3

x x 3 x x

x

27.
L’HOSPITAL’S RULE Example 4

tan x x

Find lim 3

x 0 x

Noting that both tan x – x → 0 and x3 → 0

as x → 0, we use l’Hospital’s Rule:

2

tan x x sec x 1

lim 3

lim

x 0 x x 0 3x 2

tan x x

Find lim 3

x 0 x

Noting that both tan x – x → 0 and x3 → 0

as x → 0, we use l’Hospital’s Rule:

2

tan x x sec x 1

lim 3

lim

x 0 x x 0 3x 2

28.
L’HOSPITAL’S RULE Example 4

As the limit on the right side is still

0

indeterminate of type , we apply the rule

0

again:

2 2

sec x 1 2sec x tan x

lim 2

lim

x 0 3x x 0 6x

As the limit on the right side is still

0

indeterminate of type , we apply the rule

0

again:

2 2

sec x 1 2sec x tan x

lim 2

lim

x 0 3x x 0 6x

29.
L’HOSPITAL’S RULE Example 4

2

Since lim sec x 1 , we simplify the

x 0

calculation by writing:

2

2sec x tan x 1 2 tan x

lim lim sec x lim

x 0 6x 3 x 0 x 0 x

1 tan x

lim

3 x 0 x

2

Since lim sec x 1 , we simplify the

x 0

calculation by writing:

2

2sec x tan x 1 2 tan x

lim lim sec x lim

x 0 6x 3 x 0 x 0 x

1 tan x

lim

3 x 0 x

30.
L’HOSPITAL’S RULE Example 4

We can evaluate this last limit either by

using l’Hospital’s Rule a third time or by

writing tan x as (sin x)/(cos x) and making

use of our knowledge of trigonometric limits.

We can evaluate this last limit either by

using l’Hospital’s Rule a third time or by

writing tan x as (sin x)/(cos x) and making

use of our knowledge of trigonometric limits.

31.
L’HOSPITAL’S RULE Example 4

Putting together all the steps,

we get:

2

tan x x sec x 1

lim 3

lim 2

x 0 x x 0 3x

2

2sec x tan x

lim

x 0 6x

2

1 tan x 1 sec x 1

lim lim

3 x 0 x 3 x 0 1 3

Putting together all the steps,

we get:

2

tan x x sec x 1

lim 3

lim 2

x 0 x x 0 3x

2

2sec x tan x

lim

x 0 6x

2

1 tan x 1 sec x 1

lim lim

3 x 0 x 3 x 0 1 3

32.
L’HOSPITAL’S RULE Example 5

sin x

Find xlim

1 cos x

If we blindly attempted to use l-Hospital’s rule,

we would get: sin x cos x

lim lim

x 1 cos x x sin x

sin x

Find xlim

1 cos x

If we blindly attempted to use l-Hospital’s rule,

we would get: sin x cos x

lim lim

x 1 cos x x sin x

33.
L’HOSPITAL’S RULE Example 5

This is wrong.

Although the numerator sin x → 0 as x → π -,

notice that the denominator (1 - cos x) does not

approach 0.

So, the rule can’t be applied here.

This is wrong.

Although the numerator sin x → 0 as x → π -,

notice that the denominator (1 - cos x) does not

approach 0.

So, the rule can’t be applied here.

34.
L’HOSPITAL’S RULE Example 5

The required limit is, in fact, easy to find

because the function is continuous at π

and the denominator is nonzero there:

sin x sin 0

lim 0

x 1 cos x 1 cos 1 ( 1)

The required limit is, in fact, easy to find

because the function is continuous at π

and the denominator is nonzero there:

sin x sin 0

lim 0

x 1 cos x 1 cos 1 ( 1)

35.
L’HOSPITAL’S RULE

The example shows what can go wrong

if you use the rule without thinking.

Other limits can be found using the rule, but

are more easily found by other methods.

See Examples 3 and 5 in Section 2.3,

Example 3 in Section 2.6, and the discussion

at the beginning of the section.

The example shows what can go wrong

if you use the rule without thinking.

Other limits can be found using the rule, but

are more easily found by other methods.

See Examples 3 and 5 in Section 2.3,

Example 3 in Section 2.6, and the discussion

at the beginning of the section.

36.
L’HOSPITAL’S RULE

So, when evaluating any limit,

you should consider other methods

before using l’Hospital’s Rule.

So, when evaluating any limit,

you should consider other methods

before using l’Hospital’s Rule.

37.
INDETERMINATE PRODUCTS

If lim f ( x) 0 and lim g ( x) (or -∞),

x a x a

then it isn’t clear what the value

of lim f ( x) g ( x), if any, will be.

x a

If lim f ( x) 0 and lim g ( x) (or -∞),

x a x a

then it isn’t clear what the value

of lim f ( x) g ( x), if any, will be.

x a

38.
INDETERMINATE PRODUCTS

There is a struggle between f and g.

If f wins, the answer will be 0.

If g wins, the answer will be ∞ (or -∞).

Alternatively, there may be a compromise

where the answer is a finite nonzero number.

There is a struggle between f and g.

If f wins, the answer will be 0.

If g wins, the answer will be ∞ (or -∞).

Alternatively, there may be a compromise

where the answer is a finite nonzero number.

39.
INDETERMINATE FORM—TYPE 0 . ∞

This kind of limit is called an indeterminate

form of type 0 . ∞.

We can deal with it by writing the product fg

as a quotient: f g

fg or fg

1/ g 1/ f

This converts the given limit into an indeterminate form

0

of type or ∞/∞, so that we can use l’Hospital’s Rule.

0

This kind of limit is called an indeterminate

form of type 0 . ∞.

We can deal with it by writing the product fg

as a quotient: f g

fg or fg

1/ g 1/ f

This converts the given limit into an indeterminate form

0

of type or ∞/∞, so that we can use l’Hospital’s Rule.

0

40.
INDETERMINATE PRODUCTS Example 6

Evaluate lim x ln x

x 0

The given limit is indeterminate because,

as x → 0+, the first factor (x) approaches 0,

whereas the second factor (ln x) approaches -∞.

Evaluate lim x ln x

x 0

The given limit is indeterminate because,

as x → 0+, the first factor (x) approaches 0,

whereas the second factor (ln x) approaches -∞.

41.
INDETERMINATE PRODUCTS Example 6

Writing x = 1/(1/x), we have 1/x → ∞

as x → 0+.

So, l’Hospital’s Rule gives:

ln x 1/ x

lim x ln x lim lim 2

x 0 x 0 1/ x x 0 1/ x

lim ( x) 0

x 0

Writing x = 1/(1/x), we have 1/x → ∞

as x → 0+.

So, l’Hospital’s Rule gives:

ln x 1/ x

lim x ln x lim lim 2

x 0 x 0 1/ x x 0 1/ x

lim ( x) 0

x 0

42.
INDETERMINATE PRODUCTS Note

In solving the example, another possible

option would have been to write:

x

lim x ln x lim

x 0 x 0 1/ ln x

This gives an indeterminate form of the type 0/0.

However, if we apply l’Hospital’s Rule, we get a more

complicated expression than the one we started with.

In solving the example, another possible

option would have been to write:

x

lim x ln x lim

x 0 x 0 1/ ln x

This gives an indeterminate form of the type 0/0.

However, if we apply l’Hospital’s Rule, we get a more

complicated expression than the one we started with.

43.
INDETERMINATE PRODUCTS Note

In general, when we rewrite an

indeterminate product, we try to choose

the option that leads to the simpler limit.

In general, when we rewrite an

indeterminate product, we try to choose

the option that leads to the simpler limit.

44.
INDETERMINATE FORM—TYPE ∞ -∞

If lim f ( x) and lim g ( x) , then

x a x a

the limit

lim[ f ( x) g ( x)]

x a

is called an indeterminate form

of type ∞ - ∞.

If lim f ( x) and lim g ( x) , then

x a x a

the limit

lim[ f ( x) g ( x)]

x a

is called an indeterminate form

of type ∞ - ∞.

45.
INDETERMINATE DIFFERENCES

Again, there is a contest between f and g.

Will the answer be ∞ (f wins)?

Will it be - ∞ (g wins)?

Will they compromise on a finite number?

Again, there is a contest between f and g.

Will the answer be ∞ (f wins)?

Will it be - ∞ (g wins)?

Will they compromise on a finite number?

46.
INDETERMINATE DIFFERENCES

To find out, we try to convert the difference

into a quotient (for instance, by using a

common denominator, rationalization, or

factoring out a common factor) so that

0

we have an indeterminate form of type

0

or ∞/∞.

To find out, we try to convert the difference

into a quotient (for instance, by using a

common denominator, rationalization, or

factoring out a common factor) so that

0

we have an indeterminate form of type

0

or ∞/∞.

47.
INDETERMINATE DIFFERENCES Example 7

Compute lim (sec x tan x)

x ( / 2)

First, notice that sec x → ∞ and tan x → ∞

as x → (π/2)-.

So, the limit is indeterminate.

Compute lim (sec x tan x)

x ( / 2)

First, notice that sec x → ∞ and tan x → ∞

as x → (π/2)-.

So, the limit is indeterminate.

48.
INDETERMINATE DIFFERENCES Example 7

Here, we use a common denominator:

1 sin x

lim (sec x tan x) lim

x ( / 2) x ( / 2) cos x cos x

1 sin x

lim

x ( / 2) cos x

cos x

lim 0

x ( / 2) sin x

Note that the use of l’Hospital’s Rule is justified

because 1 – sin x → 0 and cos x → 0 as x → (π/2)-.

Here, we use a common denominator:

1 sin x

lim (sec x tan x) lim

x ( / 2) x ( / 2) cos x cos x

1 sin x

lim

x ( / 2) cos x

cos x

lim 0

x ( / 2) sin x

Note that the use of l’Hospital’s Rule is justified

because 1 – sin x → 0 and cos x → 0 as x → (π/2)-.

49.
INDETERMINATE POWERS

Several indeterminate forms arise from

g ( x)

the limit lim[ f ( x)]

x a

0

1. lim f ( x) 0 and lim g ( x) 0 type 0

x a x a

0

2. lim f ( x) and lim g ( x) 0 type

x a x a

3. lim f ( x) 1 and lim g ( x) type1

x a x a

Several indeterminate forms arise from

g ( x)

the limit lim[ f ( x)]

x a

0

1. lim f ( x) 0 and lim g ( x) 0 type 0

x a x a

0

2. lim f ( x) and lim g ( x) 0 type

x a x a

3. lim f ( x) 1 and lim g ( x) type1

x a x a

50.
INDETERMINATE POWERS

Each of these three cases can be treated

in either of two ways.

Taking the natural logarithm:

Let y [ f ( x)]g ( x ) , then ln y g ( x) ln f ( x)

Writing the function as an exponential:

g ( x) g ( x )ln f ( x )

[ f ( x)] e

Each of these three cases can be treated

in either of two ways.

Taking the natural logarithm:

Let y [ f ( x)]g ( x ) , then ln y g ( x) ln f ( x)

Writing the function as an exponential:

g ( x) g ( x )ln f ( x )

[ f ( x)] e

51.
INDETERMINATE POWERS

Recall that both these methods were used

in differentiating such functions.

In either method, we are led to the indeterminate

product g(x) ln f(x), which is of type 0 . ∞.

Recall that both these methods were used

in differentiating such functions.

In either method, we are led to the indeterminate

product g(x) ln f(x), which is of type 0 . ∞.

52.
INDETERMINATE POWERS Example 8

cot x

Calculate lim (1 sin 4 x)

x 0

First, notice that, as x → 0+, we have

1 + sin 4x → 1 and cot x → ∞.

So, the given limit is indeterminate.

cot x

Calculate lim (1 sin 4 x)

x 0

First, notice that, as x → 0+, we have

1 + sin 4x → 1 and cot x → ∞.

So, the given limit is indeterminate.

53.
INDETERMINATE POWERS Example 8

Let y = (1 + sin 4x)cot x

Then, ln y = ln[(1 + sin 4x)cot x]

= cot x ln(1 + sin 4x)

Let y = (1 + sin 4x)cot x

Then, ln y = ln[(1 + sin 4x)cot x]

= cot x ln(1 + sin 4x)

54.
INDETERMINATE POWERS Example 8

So, l’Hospital’s Rule gives:

ln(1 sin 4 x)

lim ln y lim

x 0 x 0 tan x

4 cos 4 x

lim 1 sin2 4 x 4

x 0 sec x

So, l’Hospital’s Rule gives:

ln(1 sin 4 x)

lim ln y lim

x 0 x 0 tan x

4 cos 4 x

lim 1 sin2 4 x 4

x 0 sec x

55.
INDETERMINATE POWERS Example 8

So far, we have computed the limit of ln y.

However, what we want is the limit of y.

To find this, we use the fact that y = eln y:

cot x

lim (1 sin 4 x) lim y

x 0 x 0

lim eln y e 4

x 0

So far, we have computed the limit of ln y.

However, what we want is the limit of y.

To find this, we use the fact that y = eln y:

cot x

lim (1 sin 4 x) lim y

x 0 x 0

lim eln y e 4

x 0

56.
INDETERMINATE POWERS Example 9

x

Find lim x

x 0

Notice that this limit is indeterminate

since 0x = 0 for any x > 0 but x0 = 1

for any x ≠ 0.

x

Find lim x

x 0

Notice that this limit is indeterminate

since 0x = 0 for any x > 0 but x0 = 1

for any x ≠ 0.

57.
INDETERMINATE POWERS Example 9

We could proceed as in Example 8 or by

writing the function as an exponential:

xx = (eln x)x = ex ln x

In Example 6, we used l’Hospital’s Rule

to show that lim x ln x 0

x 0

x x ln x 0

Therefore, lim x lim e e 1

x 0 x 0

We could proceed as in Example 8 or by

writing the function as an exponential:

xx = (eln x)x = ex ln x

In Example 6, we used l’Hospital’s Rule

to show that lim x ln x 0

x 0

x x ln x 0

Therefore, lim x lim e e 1

x 0 x 0