L’Hopital’s Rule 31.L’Hopital ... - Auburn University

L¡¯Hopital¡¯s Rule

31. L¡¯Hopital¡¯s Rule

Limit of indeterminate type

L¡¯Ho?pital¡¯s rule

Common mistakes

31.1. Limit of indeterminate type

Examples

Some limits for which the substitution rule does not apply can be found by using inspection.

For instance,





cos x

about 1

lim

x¡ú0 x2

small pos.

Indeterminate product

Indeterminate difference

Indeterminate powers

Summary

=¡Þ

On the other hand, we have seen (8) that inspection cannot be used to find the limit of a

fraction when both numerator and denominator go to 0. The examples given were

2

lim

x¡ú0+

x

,

x

lim

x¡ú0+

x

,

x2

lim

x¡ú0+

x

.

x

In each case, both numerator and denominator go to 0. If we had a way to use inspection to

decide the limit in this case, then it would have to give the same answer in all three cases.

Yet, the first limit is 0, the second is ¡Þ and the third is 1 (as can be seen by canceling x¡¯s).

We say that each of the above limits is indeterminate of type 00 . A useful way to

remember that one cannot use inspection in this case is to imagine that the numerator

going to 0 is trying to make the fraction small, while the denominator going to 0 is trying

to make the fraction large. There is a struggle going on. In the first case above, the

numerator wins (limit is 0); in the second case, the denominator wins (limit is ¡Þ); in the

third case, there is a compromise (limit is 1).

Changing the limits above so that x goes to infinity instead gives a different indeterminate

Table of Contents

JJ

II

J

I

Page 1 of 17

Back

Print Version

Home Page

L¡¯Hopital¡¯s Rule

type. In each of the limits

Limit of indeterminate type

x2

lim

,

x¡ú¡Þ x

x

lim

,

x¡ú¡Þ x2

x

lim .

x¡ú¡Þ x

both numerator and denominator go to infinity. The numerator going to infinity is trying

to make the fraction large, while the denominator going to infinity is trying to make the

fraction small. Again, there is a struggle. Once again, we can cancel x¡¯s to see that the

first limit is ¡Þ (numerator wins), the second limit is 0 (denominator wins), and the third

limit is 1 (compromise). The different answers show that one cannot use inspection in this

¡Þ

.

case. Each of these limits is indeterminate of type ¡Þ

Sometimes limits of indeterminate types 00 or ¡Þ

¡Þ can be determined by using some algebraic

technique, like canceling between numerator and denominator as we did above (see also

12). Usually, though, no such algebraic technique suggests itself, as is the case for the limit

L¡¯Ho?pital¡¯s rule

Common mistakes

Examples

Indeterminate product

Indeterminate difference

Indeterminate powers

Summary

Table of Contents

x2

,

x¡ú0 sin x

JJ

II

which is indeterminate of type 00 . Fortunately, there is a general rule that can be applied,

namely, l¡¯Ho?pital¡¯s rule.

J

I

lim

Page 2 of 17

Back

Print Version

Home Page

L¡¯Hopital¡¯s Rule

31.2. L¡¯Ho?pital¡¯s rule

Limit of indeterminate type

L¡¯Ho?pital¡¯s rule

Common mistakes

L¡¯Ho?pital¡¯s rule. If the limit

Examples

lim

Indeterminate product

f (x)

g(x)

Indeterminate difference

Indeterminate powers

is of indeterminate type

0

0

or

lim

¡À¡Þ

¡À¡Þ ,

then

Summary

f (x)

f 0 (x)

= lim 0

,

g(x)

g (x)

provided this last limit exists. Here, lim stands for lim , lim¡À , or lim .

x¡úa x¡úa

x¡ú¡À¡Þ

The pronunciation is lo?-pe?-ta?l. Evidently, this result is actually due to the mathematician

Bernoulli rather than to l¡¯Ho?pital. The verification of l¡¯Ho?pital¡¯s rule (omitted) depends

on the mean value theorem.

31.2.1

Example

x2

.

x¡ú0 sin x

Find lim

Solution As observed above, this limit is of indeterminate type

applies. We have

 

x2

0 l¡¯H

2x

0

lim

= lim

= = 0,

x¡ú0 sin x

x¡ú0 cos x

0

1

Table of Contents

JJ

II

J

I

Page 3 of 17

0

0,

where we have first used l¡¯Ho?pital¡¯s rule and then the substitution rule.

so l¡¯Ho?pital¡¯s rule

Back

Print Version

Home Page

L¡¯Hopital¡¯s Rule

The solution of the previous example shows the notation we use to indicate the type of an

indeterminate limit and the subsequent use of l¡¯Ho?pital¡¯s rule.

31.2.2

Solution

Example

Limit of indeterminate type

L¡¯Ho?pital¡¯s rule

Common mistakes

3x ? 2

Find lim

.

2

x¡ú?¡Þ ex

Examples

Indeterminate product

Indeterminate difference

We have

3x ? 2

lim

2

x¡ú?¡Þ ex



?¡Þ

¡Þ



3

= lim x2

x¡ú?¡Þ e (2x)

= 0.

l¡¯H



3

large neg.



Indeterminate powers

Summary

Table of Contents

31.3. Common mistakes

Here are two pitfalls to avoid:

? L¡¯Ho?pital¡¯s rule should not be used if the limit is not indeterminate (of the appropriate

type). For instance, the following limit is not indeterminate; in fact, the substitution

rule applies to give the limit:

sin x

0

= = 0.

x¡ú0 x + 1

1

lim

II

J

I

Page 4 of 17

Back

An application of l¡¯Ho?pital¡¯s rule gives the wrong answer:

sin x l¡¯H

cos x

1

lim

= lim

= =1

x¡ú0 x + 1

x¡ú0

1

1

JJ

(wrong).

Print Version

Home Page

L¡¯Hopital¡¯s Rule

? Although l¡¯Ho?pital¡¯s rule involves a quotient f (x)/g(x) as well as derivatives, the

quotient rule of differentiation is not involved. The expression in l¡¯Ho?pital¡¯s rule is



0

f 0 (x)

f (x)

and not

.

g 0 (x)

g(x)

Limit of indeterminate type

L¡¯Ho?pital¡¯s rule

Common mistakes

Examples

Indeterminate product

Indeterminate difference

31.4. Examples

Indeterminate powers

Summary

31.4.1

Solution

Example

We have

sin ¦È

Find lim

.

¦È¡ú0 ¦È

sin ¦È

¦È¡ú0 ¦È

lim

 

0 l¡¯H

cos ¦È

1

= lim

= = 1.

¦È¡ú0 1

0

1

(In 19, we had to work pretty hard to determine this important limit. It is tempting to go

back and replace that argument with this much easier one, but unfortunately we used this

limit to derive the formula for the derivative of sin ¦È, which is used here in the application

of l¡¯Ho?pital¡¯s rule, so that would make for a circular argument.)

Sometimes repeated use of l¡¯Ho?pital¡¯s rule is called for:

31.4.2

Solution

Example

3x2 + x + 4

.

x¡ú¡Þ 5x2 + 8x

Find lim

We have

Table of Contents

JJ

II

J

I

Page 5 of 17

Back

Print Version

2

3x + x + 4  ¡Þ  l¡¯H

6x + 1  ¡Þ  l¡¯H

6

6

3

lim

=

lim

= lim

=

= .

x¡ú¡Þ 5x2 + 8x

x¡ú¡Þ 10x + 8

x¡ú¡Þ 10

¡Þ

¡Þ

10

5

Home Page

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download