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## 6.8 Indeterminate Forms and

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### Indeterminate Forms and l’Hospital’s Rule

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. In particular, we would like to know the value of the limit

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### Indeterminate Forms and l’Hospital’s Rule

In computing this limit we can’t apply Law 5 of limits

because the limit of the denominator is 0. In fact, although the limit in (1) exists, its value is not obvious because both numerator and denominator approach 0 and is not

defined.

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 and is called an indeterminate form

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### Indeterminate Forms and l’Hospital’s Rule

For rational functions, we can cancel common factors:

We used a geometric argument to show that

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### Indeterminate Forms and l’Hospital’s Rule

But these methods do not work for limits such as (1), so in this section we introduce a systematic method, known as l’Hospital’s Rule, for the evaluation of 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

It isn’t obvious how to evaluate this limit because both

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### Indeterminate Forms and l’Hospital’s Rule

There is a struggle between numerator and denominator. If the numerator wins, the limit will be ; if the denominator wins, the answer will be 0. Or there may be some

compromise, in which case the answer will be some finite positive number.

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 and is called an

indeterminate form of type / .

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### Indeterminate Forms and l’Hospital’s Rule

This type of limit can be evaluated for certain functions, including rational functions, by dividing numerator and denominator by the highest power of x that occurs in the denominator. For instance,

This method does not work for limits such as (2).

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### Indeterminate Forms and l’Hospital’s Rule

L’Hospital’s Rule also applies to this type of indeterminate form.

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### Indeterminate Forms and l’Hospital’s Rule

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 l’Hospital’s Rule.

Note 2:

L’Hospital’s 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 → – .

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### Indeterminate Forms and l’Hospital’s Rule

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 l’Hospital’s Rule is true. In fact, using the alternative form of the

definition of a derivative, we have

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### Indeterminate Forms and l’Hospital’s Rule

The general version of l’Hospital’s Rule for the

indeterminate form is somewhat more difficult to prove.

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### Example 1

Find

Solution:

Since

and

the limit is an indeterminate form of type .

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### Example 1 – Solution

We can apply l’Hospital’s Rule:

cont’d

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### Indeterminate Products

If limx→a f(x) = 0 and limx→a g(x) = (or – ), then it isn’t clear what the value of limx→a [f(x)g(x)], if any, will be.

There is a struggle between f and g. If f wins, the answer will be 0; if g wins, the answer will be (or – ).

Or 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 .

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### Indeterminate Products

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.

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### Example 6

Evaluate

Solution:

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

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### Example 6 – Solution

Writing x = 1/(1/x), we have 1/xas x → 0+, so l’Hospital’s Rule gives

= 0

cont’d

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### Indeterminate Products

Note:

In solving Example 6 another possible option would have been to write

This gives an indeterminate form of the type , but if we apply l’Hospital’s Rule we get a more complicated

expression than the one we started with.

In general, when we rewrite an indeterminate product, we try to choose the option that leads to the simpler limit.

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### Indeterminate Differences

If limx→a f(x) = and limx→a g(x) = , then the limit

is called an indeterminate form of type – . Again there is a contest between f and g.

Will the answer be (f wins) or will it be – (g wins) or 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, or rationalization, or factoring out

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### Example 8

Compute

Solution:

First notice that sec xand tan xas x → (π/2), so the limit is indeterminate.

Here we use a common denominator:

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### Example 8 – Solution

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

cont’d

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### Indeterminate Powers

Several indeterminate forms arise from the limit

1. and type 00 2. and type 0 3. and type

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### Indeterminate Powers

Each of these three cases can be treated either by taking the natural logarithm:

let y = [f(x)]g(x), then ln y = g(x) ln f(x) or by writing the function as an exponential:

[f(x)]g(x) = eg(x) ln f(x)

In either method we are led to the indeterminate product g(x) ln f(x), which is of type 0 .

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### Example 9

Calculate

Solution:

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

cot x → , so the given limit is indeterminate (type ). Let y = (1 + sin 4x)cot x

Then

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

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### Example 9 – Solution

So l’Hospital’s Rule gives

So far we have computed the limit of ln y, but what we want is the limit of y.

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

cont’d

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### Indeterminate Powers

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