31
Learning Objectives
- Recognize when to apply L’Hôpital’s rule.
- Identify indeterminate forms produced by quotients, products, subtractions, and powers, and apply L’Hôpital’s rule in each case.
- Describe the relative growth rates of functions.
In this section, we examine a powerful tool for evaluating limits. This tool, known as L’Hôpital’s rule , uses derivatives to calculate limits. With this rule, we will be able to evaluate many limits we have not yet been able to determine. Instead of relying on numerical evidence to conjecture that a limit exists, we will be able to show definitively that a limit exists and to determine its exact value.
Applying L’Hôpital’s Rule
L’Hôpital’s rule can be used to evaluate limits involving the quotient of two functions. Consider

If then

However, what happens if and
We call this one of the indeterminate forms , of type
This is considered an indeterminate form because we cannot determine the exact behavior of
as
without further analysis. We have seen examples of this earlier in the text. For example, consider

For the first of these examples, we can evaluate the limit by factoring the numerator and writing

For we were able to show, using a geometric argument, that

Here we use a different technique for evaluating limits such as these. Not only does this technique provide an easier way to evaluate these limits, but also, and more important, it provides us with a way to evaluate many other limits that we could not calculate previously.
The idea behind L’Hôpital’s rule can be explained using local linear approximations. Consider two differentiable functions and
such that
and such that
For
near
we can write

and

Therefore,





Since is differentiable at
then
is continuous at
and therefore
Similarly,
If we also assume that
and
are continuous at
then
and
Using these ideas, we conclude that

Note that the assumption that and
are continuous at
and
can be loosened. We state L’Hôpital’s rule formally for the indeterminate form
Also note that the notation
does not mean we are actually dividing zero by zero. Rather, we are using the notation
to represent a quotient of limits, each of which is zero.
L’Hôpital’s Rule (0/0 Case)
Suppose and
are differentiable functions over an open interval containing
except possibly at
If
and
then

assuming the limit on the right exists or is or
This result also holds if we are considering one-sided limits, or if
Proof
We provide a proof of this theorem in the special case when and
are all continuous over an open interval containing
In that case, since
and
and
are continuous at
it follows that
Therefore,

Note that L’Hôpital’s rule states we can calculate the limit of a quotient by considering the limit of the quotient of the derivatives
It is important to realize that we are not calculating the derivative of the quotient
□
Applying L’Hôpital’s Rule (0/0 Case)
Evaluate each of the following limits by applying L’Hôpital’s rule.
Solution
- Since the numerator
and the denominator
we can apply L’Hôpital’s rule to evaluate this limit. We have
- As
the numerator
and the denominator
Therefore, we can apply L’Hôpital’s rule. We obtain
- As
the numerator
and the denominator
Therefore, we can apply L’Hôpital’s rule. We obtain
- As
both the numerator and denominator approach zero. Therefore, we can apply L’Hôpital’s rule. We obtain
Since the numerator and denominator of this new quotient both approach zero as
we apply L’Hôpital’s rule again. In doing so, we see that
Therefore, we conclude that
Evaluate
Hint
Solution
1
We can also use L’Hôpital’s rule to evaluate limits of quotients in which
and
Limits of this form are classified as indeterminate forms of type
Again, note that we are not actually dividing
by
Since
is not a real number, that is impossible; rather,
is used to represent a quotient of limits, each of which is
or
L’Hôpital’s Rule
Case)
Suppose and
are differentiable functions over an open interval containing
except possibly at
Suppose
(or
and
(or
Then,

assuming the limit on the right exists or is or
This result also holds if the limit is infinite, if
or
or the limit is one-sided.
Applying L’Hôpital’s Rule
Case)
Evaluate each of the following limits by applying L’Hôpital’s rule.
Solution
- Since
and
are first-degree polynomials with positive leading coefficients,
and
Therefore, we apply L’Hôpital’s rule and obtain
Note that this limit can also be calculated without invoking L’Hôpital’s rule. Earlier in the chapter we showed how to evaluate such a limit by dividing the numerator and denominator by the highest power of
in the denominator. In doing so, we saw that
L’Hôpital’s rule provides us with an alternative means of evaluating this type of limit.
- Here,
and
Therefore, we can apply L’Hôpital’s rule and obtain
Now as
Therefore, the first term in the denominator is approaching zero and the second term is getting really large. In such a case, anything can happen with the product. Therefore, we cannot make any conclusion yet. To evaluate the limit, we use the definition of
to write
Now
and
so we apply L’Hôpital’s rule again. We find
We conclude that
Evaluate
Hint
Solution
0
As mentioned, L’Hôpital’s rule is an extremely useful tool for evaluating limits. It is important to remember, however, that to apply L’Hôpital’s rule to a quotient it is essential that the limit of
be of the form
or
Consider the following example.
When L’Hôpital’s Rule Does Not Apply
Consider Show that the limit cannot be evaluated by applying L’Hôpital’s rule.
Solution
Because the limits of the numerator and denominator are not both zero and are not both infinite, we cannot apply L’Hôpital’s rule. If we try to do so, we get

and

At which point we would conclude erroneously that

However, since and
we actually have

We can conclude that

When L’Hôpital’s Rule Does Not Apply
Consider Show that the limit cannot be evaluated by applying L’Hôpital’s rule.
Solution
The limits of the numerator and denominator are both zero. Thus it appears that we can apply L’Hôpital’s rule. However, when we try to do so, we get

and

At which point we would conclude erroneously that

However, since neither exists, nor is infinite, L’Hôpital’s rule does not apply.
However, this does not mean the limit fails to exist. Using the squeeze theorem, we get that the function is bounded between
and
, both of whose limits are
We can conclude that the by squeeze theorem

Explain why we cannot apply L’Hôpital’s rule to evaluate Evaluate
by other means.
Hint
Determine the limits of the numerator and denominator separately.
Solution
Therefore, we cannot apply L’Hôpital’s rule. The limit of the quotient is
Other Indeterminate Forms
L’Hôpital’s rule is very useful for evaluating limits involving the indeterminate forms and
However, we can also use L’Hôpital’s rule to help evaluate limits involving other indeterminate forms that arise when evaluating limits. The expressions
and
are all considered indeterminate forms. These expressions are not real numbers. Rather, they represent forms that arise when trying to evaluate certain limits. Next we realize why these are indeterminate forms and then understand how to use L’Hôpital’s rule in these cases. The key idea is that we must rewrite the indeterminate forms in such a way that we arrive at the indeterminate form
or
Indeterminate Form of Type 
Suppose we want to evaluate where
and
(or
as
Since one term in the product is approaching zero but the other term is becoming arbitrarily large (in magnitude), anything can happen to the product. We use the notation
to denote the form that arises in this situation. The expression
is considered indeterminate because we cannot determine without further analysis the exact behavior of the product
as
For example, let
be a positive integer and consider

As
and
However, the limit as
of
varies, depending on
If
then
If
then
If
then
Here we consider another limit involving the indeterminate form
and show how to rewrite the function as a quotient to use L’Hôpital’s rule.
Indeterminate Form of Type 
Evaluate
Solution
First, rewrite the function as a quotient to apply L’Hôpital’s rule. If we write

we see that as
and
as
Therefore, we can apply L’Hôpital’s rule and obtain

We conclude that




Evaluate
Hint
Write
Solution
1
Indeterminate Form of Type 
Another type of indeterminate form is Consider the following example. Let
be a positive integer and let
and
As
and
We are interested in
Depending on whether
grows faster,
grows faster, or they grow at the same rate, as we see next, anything can happen in this limit. Since
and
we write
to denote the form of this limit. As with our other indeterminate forms,
has no meaning on its own and we must do more analysis to determine the value of the limit. For example, suppose the exponent
in the function
is
then

On the other hand, if then

However, if then

Therefore, the limit cannot be determined by considering only Next we see how to rewrite an expression involving the indeterminate form
as a fraction to apply L’Hôpital’s rule.
Indeterminate Form of Type 
Evaluate
Solution
By combining the fractions, we can write the function as a quotient. Since the least common denominator is we have

As the numerator
and the denominator
Therefore, we can apply L’Hôpital’s rule. Taking the derivatives of the numerator and the denominator, we have

As
and
Since the denominator is positive as
approaches zero from the right, we conclude that

Therefore,

Evaluate
Hint
Rewrite the difference of fractions as a single fraction.
Solution
0
Another type of indeterminate form that arises when evaluating limits involves exponents. The expressions
and
are all indeterminate forms. On their own, these expressions are meaningless because we cannot actually evaluate these expressions as we would evaluate an expression involving real numbers. Rather, these expressions represent forms that arise when finding limits. Now we examine how L’Hôpital’s rule can be used to evaluate limits involving these indeterminate forms.
Since L’Hôpital’s rule applies to quotients, we use the natural logarithm function and its properties to reduce a problem evaluating a limit involving exponents to a related problem involving a limit of a quotient. For example, suppose we want to evaluate and we arrive at the indeterminate form
(The indeterminate forms
and
can be handled similarly.) We proceed as follows. Let

Then,

Therefore,
![Rendered by QuickLaTeX.com \underset{x\to a}{\text{lim}}\left[\text{ln}(y)\right]=\underset{x\to a}{\text{lim}}\left[g(x)\text{ln}(f(x))\right].](https://opentextbooks.clemson.edu/app/uploads/quicklatex/quicklatex.com-d54d0894b6a9ea254acfb0bc668e6d6a_l3.png)
Since we know that
Therefore,
is of the indeterminate form
and we can use the techniques discussed earlier to rewrite the expression
in a form so that we can apply L’Hôpital’s rule. Suppose
where
may be
or
Then
![Rendered by QuickLaTeX.com \underset{x\to a}{\text{lim}}\left[\text{ln}(y)\right]=L.](https://opentextbooks.clemson.edu/app/uploads/quicklatex/quicklatex.com-65b5322c7b8aaf42158d5d386147149b_l3.png)
Since the natural logarithm function is continuous, we conclude that

which gives us

Indeterminate Form of Type 
Evaluate
Solution
Let Then,

We need to evaluate Applying L’Hôpital’s rule, we obtain

Therefore, Since the natural logarithm function is continuous, we conclude that

which leads to

Hence,

Evaluate
Hint
Let and apply the natural logarithm to both sides of the equation.
Solution
Indeterminate Form of Type 
Evaluate
Solution
Let

Therefore,

We now evaluate Since
and
we have the indeterminate form
To apply L’Hôpital’s rule, we need to rewrite
as a fraction. We could write

or

Let’s consider the first option. In this case, applying L’Hôpital’s rule, we would obtain

Unfortunately, we not only have another expression involving the indeterminate form but the new limit is even more complicated to evaluate than the one with which we started. Instead, we try the second option. By writing

and applying L’Hôpital’s rule, we obtain

Using the fact that and
we can rewrite the expression on the right-hand side as
![Rendered by QuickLaTeX.com \underset{x\to {0}^{+}}{\text{lim}}\frac{-{ \sin }^{2}x}{x \cos x}=\underset{x\to {0}^{+}}{\text{lim}}\left[\frac{ \sin x}{x}\cdot(- \tan x)\right]=(\underset{x\to {0}^{+}}{\text{lim}}\frac{ \sin x}{x})\cdot(\underset{x\to {0}^{+}}{\text{lim}}(- \tan x))=1\cdot0=0.](https://opentextbooks.clemson.edu/app/uploads/quicklatex/quicklatex.com-d14ccce9d400eeabe576662a7ba75da2_l3.png)
We conclude that Therefore,
and we have

Hence,

Evaluate
Hint
Let and take the natural logarithm of both sides of the equation.
Solution
1
Growth Rates of Functions
Suppose the functions and
both approach infinity as
Although the values of both functions become arbitrarily large as the values of
become sufficiently large, sometimes one function is growing more quickly than the other. For example,
and
both approach infinity as
However, as shown in the following table, the values of
are growing much faster than the values of
![]() |
10 | 100 | 1000 | 10,000 |
![]() |
100 | 10,000 | 1,000,000 | 100,000,000 |
![]() |
1000 | 1,000,000 | 1,000,000,000 | ![]() |
In fact,

As a result, we say is growing more rapidly than
as
On the other hand, for
and
although the values of
are always greater than the values of
for
each value of
is roughly three times the corresponding value of
as
as shown in the following table. In fact,

![]() |
10 | 100 | 1000 | 10,000 |
![]() |
100 | 10,000 | 1,000,000 | 100,000,000 |
![]() |
341 | 30,401 | 3,004,001 | 300,040,001 |
In this case, we say that and
are growing at the same rate as
More generally, suppose and
are two functions that approach infinity as
We say
grows more rapidly than
as
if

On the other hand, if there exists a constant such that

we say and
grow at the same rate as
Next we see how to use L’Hôpital’s rule to compare the growth rates of power, exponential, and logarithmic functions.
Comparing the Growth Rates of
and 
For each of the following pairs of functions, use L’Hôpital’s rule to evaluate
Solution
- Since
and
we can use L’Hôpital’s rule to evaluate
We obtain
Since
and
we can apply L’Hôpital’s rule again. Since
we conclude that
Therefore,
grows more rapidly than
as
(See (Figure) and (Figure) ).
Figure 3. An exponential function grows at a faster rate than a power function. Growth rates of a power function and an exponential function. 5 10 15 20 25 100 225 400 148 22,026 3,269,017 485,165,195 - Since
and
we can use L’Hôpital’s rule to evaluate
We obtain
Thus,
grows more rapidly than
as
(see (Figure) and (Figure) ).
Figure 4. A power function grows at a faster rate than a logarithmic function. Growth rates of a power function and a logarithmic function 10 100 1000 10,000 2.303 4.605 6.908 9.210 100 10,000 1,000,000 100,000,000
Compare the growth rates of and
Hint
Apply L’Hôpital’s rule to
Solution
The function grows faster than
Using the same ideas as in (Figure) a. it is not difficult to show that grows more rapidly than
for any
In (Figure) and (Figure) , we compare
with
and
as








![]() |
5 | 10 | 15 | 20 |
![]() |
125 | 1000 | 3375 | 8000 |
![]() |
625 | 10,000 | 50,625 | 160,000 |
![]() |
148 | 22,026 | 3,269,017 | 485,165,195 |
Similarly, it is not difficult to show that grows more rapidly than
for any
In (Figure) and (Figure) , we compare
with
and





![]() |
10 | 100 | 1000 | 10,000 |
![]() |
2.303 | 4.605 | 6.908 | 9.210 |
![]() |
2.154 | 4.642 | 10 | 21.544 |
![]() |
3.162 | 10 | 31.623 | 100 |
Key Concepts
- L’Hôpital’s rule can be used to evaluate the limit of a quotient when the indeterminate form
or
arises.
- L’Hôpital’s rule can also be applied to other indeterminate forms if they can be rewritten in terms of a limit involving a quotient that has the indeterminate form
or
- The exponential function
grows faster than any power function
- The logarithmic function
grows more slowly than any power function
For the following exercises, evaluate the limit.
1. Evaluate the limit
2. Evaluate the limit
Solution
3. Evaluate the limit
4. Evaluate the limit
Solution
5. Evaluate the limit
6. Evaluate the limit
Solution
For the following exercises, determine whether you can apply L’Hôpital’s rule directly. Explain why or why not. Then, indicate if there is some way you can alter the limit so you can apply L’Hôpital’s rule.
7.
8.
Solution
Cannot apply directly; use logarithms
9.
10.
Solution
Cannot apply directly; rewrite as
11.
For the following exercises, evaluate the limits with either L’Hôpital’s rule or previously learned methods.
12.
Solution
6
13.
14.
Solution
-2
15.
16.
Solution
-1
17.
18.
Solution
19.
20.
Solution
21.
22.
Solution
23.
24.
Solution
1
25.
26.
Solution
27.
28.
Solution
1
29.
30.
Solution
0
31.
32.
33.
34.
Solution
-1
35.
36.
Solution
37.
38.
Solution
1
39.
40.
Solution
For the following exercises, use a calculator to graph the function and estimate the value of the limit, then use L’Hôpital’s rule to find the limit directly.
41. [T]
42. [T]
Solution
0
43. [T]
44. [T]
45. [T]
46. [T]
Solution
0
47. [T]
48. [T]
Solution
49. [T]
50. [T]
Solution
2
Glossary
- indeterminate forms
- when evaluating a limit, the forms
and
are considered indeterminate because further analysis is required to determine whether the limit exists and, if so, what its value is
- L’Hôpital’s rule
- if
and
are differentiable functions over an interval
except possibly at
and
or
and
are infinite, then
assuming the limit on the right exists or is
or