APPLICATIONS OF DIFFERENTIATION 4. 4.4 Indeterminate Forms and L’Hospital’s Rule APPLICATIONS OF...

APPLICATIONS OF DIFFERENTIATIONAPPLICATIONS OF DIFFERENTIATION

4

4.4Indeterminate Forms

and L’Hospital’s Rule

APPLICATIONS OF DIFFERENTIATION

In this section, we will learn:

How to evaluate functions whose

values cannot be found at certain points.

Suppose we are trying to analyze

the behavior of the function

Although F is not defined when x = 1,

we need to know how F behaves near 1.

ln( )

1

xF x

x

INDETERMINATE FORMS

In particular, we would like to know

the value of the limit

1

lnlim

1x

x

x

Expression 1INDETERMINATE 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 and denominator approach 0 and is not defined.0

0

INDETERMINATE FORMS

INDETERMINATE FORM —TYPE 0/0

In general, if we have a limit of the form

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

then this limit may or may not exist.

It is called an indeterminate form of type .

We met some limits of this type in Chapter 2.

( )lim

( )x a

f x

g x

0

0

INDETERMINATE FORMS

For rational functions, we can cancel

common factors:

2

21 1

1

( 1)lim lim

( 1)( 1)1

1lim

1 2

x x

x

x x x x

x xx

x

x

INDETERMINATE FORMS

We used a geometric argument

to show that:

0

sinlim 1x

x

x

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.

INDETERMINATE FORMS

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 lnlim

1x

x

x

Expression 2

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.

INDETERMINATE FORM —TYPE ∞/∞

In general, if we have a limit of the form

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 ∞/∞.

( )lim

( )x a

f x

g x

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,2 2

2

2

11

1 1 0 1lim lim

1 2 0 22 1 2x x

x xx

x

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.

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

or that

In other words, we have an indeterminate form of type or ∞/∞.

lim ( ) 0 and lim ( ) 0x a x a

f x g x

0

0

lim ( ) and lim ( )x a x a

f x g x


Then,

if the limit on the right exists

(or is ∞ or - ∞).

( ) '( )lim lim

( ) '( )x a x a

f x f x

g x g x

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.

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 → - ∞.

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.

NOTE 3

In fact, using the alternative form of

the definition of a derivative, we have:

( ) ( ) ( ) ( )lim

'( ) '( )lim lim

( ) ( ) ( ) ( )'( ) '( ) lim

( ) ( )lim

( ) ( )

( )lim

( )

x a

x a x a

x a

x a

x a

f x f a f x f af x f a x a x a

g x g a g x g ag x g ax a x a

f x f a

g x g a

f x

g x

NOTE 3

It is more difficult to prove

the general version of l’Hospital’s

Rule.


Find

and

Thus, we can apply l’Hospital’s Rule:

1

lnlim

1x

x

x

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

x x

1 1 1 1

(ln )ln 1/ 1

lim lim lim lim 11 1( 1)

x x x x

dx

x xdxdx xxdx

Example 1


Calculate

We have and

So, l’Hospital’s Rule gives:

2lim

x

x

e

x

2lim limx

x xe x

22

( )lim lim lim

2( )

xx x

x x x

dee edx

dx xxdx

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: 2

lim lim lim2 2

x x x

x x x

e e e

x x

Example 2


Calculate

As ln x → ∞ and as x → ∞, l’Hospital’s Rule applies:

Notice that the limit on the right side is now indeterminate of type .

3

lnlimx

x

x

3 x

2/3133

ln 1/lim limx x

x x

xx

0

0

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:

2 / 313 33

ln 1/ 3lim lim lim 0x x x

x x

xx x

Example 3


Find

Noting that both tan x – x → 0 and x3 → 0 as x → 0, we use l’Hospital’s Rule:

30

tanlimx

x x

x

2

3 20 0

tan sec 1lim lim

3x x

x x x

x x

Example 4


As the limit on the right side is still indeterminate of type , we apply the rule again:

Example 4

2 2

20 0

sec 1 2sec tanlim lim

63x x

x x x

xx

0

0


Since , we simplify the calculation by writing:

2

0limsec 1x

x

22

0 0 0

0

2sec tan 1 tanlim limsec lim

6 31 tan

lim3

x x x

x

x x xx

x xx

x

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.

Example 4


Putting together all the steps,

we get:2

3 20 0

2

0

2

0 0

tan sec 1lim lim

3

2sec tanlim

6

1 tan 1 sec 1lim lim

3 3 1 3

x x

x

x x

x x x

x x

x x

x

x x

x

Example 4


Find

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

sinlim

1 cosx

x

x

sin coslim lim

1 cos sinx x

x x

x x

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.

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 sin 0lim 0

1 cos 1 cos 1 ( 1)x

x

x

Example 5


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.


So, when evaluating any limit,

you should consider other methods

before using l’Hospital’s Rule.

INDETERMINATE PRODUCTS

If and (or -∞),

then it isn’t clear what the value

of , if any, will be.

lim ( ) 0x a

f x

lim ( )

x ag x

lim ( ) ( )x a

f x g x


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.

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:

This converts the given limit into an indeterminate form of type or ∞/∞, so that we can use l’Hospital’s Rule.

or1/ 1/

f gfg fg

g f

0

0

INDETERMINATE FORM—TYPE 0 . ∞


Evaluate

The given limit is indeterminate because, as x → 0+, the first factor (x) approaches 0, whereas the second factor (ln x) approaches -∞.

0lim lnx

x x

Example 6


Writing x = 1/(1/x), we have 1/x → ∞ as x → 0+.


Example 6

20 0 0

0

ln 1/lim ln lim lim

1/ 1/lim ( ) 0

x x x

x

x xx x

x xx

In solving the example, another possible

option would have been to write:

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.

0 0lim ln lim

1/ lnx x

xx x

x

INDETERMINATE PRODUCTS Note


In general, when we rewrite an

indeterminate product, we try to choose

the option that leads to the simpler limit.

Note

If and , then

the limit

is called an indeterminate form

of type ∞ - ∞.

lim ( )x a

f x

lim ( )x ag x

lim[ ( ) ( )]x a

f x g x

INDETERMINATE FORM—TYPE ∞ - ∞

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?


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

we have an indeterminate form of type

or ∞/∞.

0

0


Compute

First, notice that sec x → ∞ and tan x → ∞ as x → (π/2)

-.

So, the limit is indeterminate.

( / 2)lim (sec tan )

xx x

Example 7


Here, we use a common denominator:

Note that the use of l’Hospital’s Rule is justified because 1 – sin x → 0 and cos x → 0 as x → (π/2)

-.

( / 2) ( / 2)

( / 2)

( / 2)

1 sinlim (sec tan ) lim

cos cos

1 sinlim

coscos

lim 0sin

x x

x

x

xx x

x x

x

xx

x

Example 7

INDETERMINATE POWERS

Several indeterminate forms arise from

the limit( )lim[ ( )]g x

x af x

0

0

1. lim ( ) 0 and lim ( ) 0 type0

2. lim ( ) and lim ( ) 0 type

3. lim ( ) 1 and lim ( ) type1

x a x a

x a x a

x a x a

f x g x

f x g x

f x g x


Each of these three cases can be treated

in either of two ways.

Taking the natural logarithm:

Writing the function as an exponential:

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

( ) ( ) ln ( )[ ( )]g x g x f xf x e


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 . ∞.


Calculate

First, notice that, as x → 0+, we have 1 + sin 4x → 1 and cot x → ∞.

So, the given limit is indeterminate.

cot

0lim (1 sin 4 ) x

xx

Example 8


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

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

= cot x ln(1 + sin 4x)

Example 8



0 0

20

ln(1 sin 4 )lim ln lim

tan4cos 4

1 sin 4lim 4sec

x x

x

xy

xx

xx

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

0 0

ln 4

0

lim(1 sin 4 ) lim

lim

x

x x

y

x

x y

e e

Example 8


Find

Notice that this limit is indeterminate since 0

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

for any x ≠ 0.

0lim x

xx

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

Therefore,

0lim ln 0x

x x

ln 0

0 0lim lim 1x x x

x xx e e

Example 9

APPLICATIONS OF DIFFERENTIATION 4. 4.4 Indeterminate Forms and L’Hospital’s Rule APPLICATIONS OF...

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