**6.8** Indeterminate Forms and

### l’Hospital’s Rule

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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

### 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

### 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** / .

### 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.

### 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

### 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 .

*Example 1 – Solution*

We can apply l’Hospital’s Rule:

cont’d

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### Indeterminate Products

### Indeterminate Products

If lim_{x→a}*f(x) = 0 and lim*_{x→a}*g(x) = (or –* ), then it isn’t
clear what the value of lim_{x→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.

### 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/x* → *as x* → 0^{+}, so
l’Hospital’s Rule gives

= 0

cont’d

### 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

### Indeterminate Differences

If lim_{x→a}*f(x) = and lim*_{x→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 x* → *and tan x* → *as x* → (π/2)^{–},
so the limit is indeterminate.

Here we use a common denominator:

*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

### Indeterminate Powers

Several indeterminate forms arise from the limit

**1.** and type 0^{0}
**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)}*= e**g(x) ln f(x)*

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

### 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 = e** ^{ln y}*:

cont’d