Introduction

Welcome to AP Calculus AB/BC's Topic 4.7 - Using L'Hospital's Rule for Determining Limits of Indeterminate Forms! In this guide, we will learn an extremely helpful way to evaluate limits.

Important Review

Recall that when evaluating limits, you should always attempt direct substitution first.

Questions

1. What is $\underset{x\to 4}{\lim }\frac{x-4}{x^2+1-5}$?

2. What is $\underset{x\to \pi }{\lim }\frac{\cos x}{x}$?

Solutions

1. 
$\underset{x\to 4}{\lim }\frac{x-4}{x^2+1-5}=\frac{(4)-4}{(4{)}^2+1-5}=\frac{0}{12}=0$

2.
$\underset{x\to \pi }{\lim }\frac{\cos x}{x}=\frac{\cos \pi }{\pi }=-\frac{1}{\pi }$

Indeterminate Forms

On the two questions above, we were able to apply direct substitution to evaluate our limits. What happens if we cannot use direct substitution?

Questions

1. If you attempt direct substitution, what does $\underset{x\to 0}{\lim }\frac{\sin (4x)}{9x}$ yield?

2. If you attempt direct substitution, what does $\underset{x\to \infty }{\lim }\frac{e^{2x}}{x^2}$ yield?

Solutions

1.
$\underset{x\to 0}{\lim }\frac{\sin (4x)}{9x}=\frac{\sin (0)}{0}=\frac{0}{0}$

2.
$\underset{x\to \infty }{\lim }\frac{e^{2x}}{x^2}=\frac{e^{2\infty }}{{\infty }^2}=\frac{\infty }{\infty }$

Note: Technically, we can never plug in $\infty$. This is for reasons that are beyond the scope of the AP Exam. Because it is beyond the scope of the AP Exam, you may plug in infinity when evaluating limits, but be careful when doing so. Your AP Calculus teacher may not like this, so ask them first. If you are confused on what this note is saying, just completely ignore it.

Notice how questions one and two have the form of either $\frac{0}{0}$ or $\frac{\infty }{\infty }$. These two forms are called indeterminate forms. These are not the only indeterminate forms, but they are the two most common.

Question

What are the two most common indeterminate forms?

Answer

$\frac{0}{0}$ and $\frac{\infty }{\infty }$

When we have indeterminate forms, we can use L'Hospital's rule.

L'Hospital's Rule

L'Hospital's rule states that if:

$\underset{x\to a}{\lim }\frac{f(x)}{g(x)}=\frac{0}{0}$ or $\frac{\infty }{\infty }$ (aka an indeterminate form)

Then:

$\underset{x\to a}{\lim }\frac{f(x)}{g(x)}=\underset{x\to a}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}$.

Let's go through some examples.

Questions

1. What is $\underset{x\to 0}{\lim }\frac{\sin x}{x}$?

2. What is $\underset{x\to \infty }{\lim }\frac{e^x}{100,000,000\cdot x}$?

Solution

1.
Using direct substitution, $\underset{x\to 0}{\lim }\frac{\sin x}{x}$ yields $\frac{0}{0}$. Therefore, we can use L'Hospital's rule. Remember that we cannot use L'Hospital's rule if we do not have $\frac{0}{0}$ or $\frac{\infty }{\infty }$ (aka an indeterminate form).

$\underset{x\to 0}{\lim }\frac{\sin x}{x}=\underset{x\to 0}{\lim }\frac{\frac{d}{dx}(\sin x)}{\frac{d}{dx}(x)}=\underset{x\to 0}{\lim }\frac{\cos x}{1}=\frac{\cos 0}{1}=1$

2.
Using direct substitution, $\underset{x\to \infty }{\lim }\frac{e^x}{100,000,000\cdot x}$ yields $\frac{\infty }{\infty }$. Therefore, we can use L'Hospital's rule. Remember that we cannot use L'Hospital's rule if we do not have $\frac{0}{0}$ or $\frac{\infty }{\infty }$ (aka an indeterminate form).

$\underset{x\to \infty }{\lim }\frac{e^x}{100,000,000\cdot x}=\underset{x\to \infty }{\lim }\frac{\frac{d}{dx}(e^x)}{\frac{d}{dx}(100,000,000\cdot x)}$

$=\underset{x\to \infty }{\lim }\frac{e^x}{100,000,000}=\frac{e^{\infty }}{100,000,000}=\infty$

With this new rule, we might be able to answer the questions from above.

Questions

1. What is $\underset{x\to 0}{\lim }\frac{\sin (4x)}{9x}$?

2. What is $\underset{x\to \infty }{\lim }\frac{e^{2x}}{x^2}$?

Solutions

Remember to always try direct substitution first!

1.
$\underset{x\to 0}{\lim }\frac{\sin (4x)}{9x}=\frac{\sin (0)}{0}=\frac{0}{0}$. Therefore, we can apply L'Hopital's rule.

$\underset{x\to 0}{\lim }\frac{\sin (4x)}{9x}=\underset{x\to 0}{\lim }\frac{\frac{d}{dx}(\sin (4x))}{\frac{d}{dx}(9x)}=\underset{x\to 0}{\lim }\frac{4\cos (4x)}{9}=\frac{4\cos (0)}{9}=\frac{4}{9}$

2.
$\underset{x\to \infty }{\lim }\frac{e^{2x}}{x^2}=\frac{e^{2\infty }}{{\infty }^2}$ $=$$\frac{\infty }{\infty }$. Therefore, we can use L'Hospital's rule.

$\underset{x\to \infty }{\lim }\frac{e^{2x}}{x^2}=\underset{x\to \infty }{\lim }\frac{\frac{d}{dx}(e^{2x})}{\frac{d}{dx}(x^2)}=\underset{x\to \infty }{\lim }\frac{2e^{2x}}{2x}=\frac{2e^{(2\infty )}}{2\infty }=\frac{\infty }{\infty }$?!?

As you can see, despite using L'Hospital's rule, we still have an indeterminate form. Because we still have this indeterminate form, we can apply L'Hospital's rule again. Remember that any time you have an indeterminate form, you can apply L'Hospital's rule. Make sure you are using L'Hospital's rule again with the new limit (the one you got when using L'Hospital's rule the first time. See the solution below to clarify what I mean).

Attempt Two of 2.
$\underset{x\to \infty }{\lim }\frac{e^{2x}}{x^2}=\frac{e^{2\infty }}{{\infty }^2}$ $=$$\frac{\infty }{\infty }$. Therefore, we can use L'Hospital's rule.

$\underset{x\to \infty }{\lim }\frac{e^{2x}}{x^2}=\underset{x\to \infty }{\lim }\frac{\frac{d}{dx}(e^{2x})}{\frac{d}{dx}(x^2)}=\underset{x\to \infty }{\lim }\frac{2e^{2x}}{2x}=\frac{2e^{(2\infty )}}{2\infty }=\frac{\infty }{\infty }$. Therefore, we can use L'Hospital's rule again.

$\underset{x\to \infty }{\lim }\frac{2e^{2x}}{2x}=\underset{x\to \infty }{\lim }\frac{\frac{d}{dx}(2e^{2x})}{\frac{d}{dx}(2x)}=\underset{x\to \infty }{\lim }\frac{4e^{2x}}{2}=\frac{4e^{(2\infty )}}{2}=\infty$!

Therefore, our final answer is the limit equals $\infty$.

Using L'Hospital's Multiple Times

We already explained this above, but we can use L'Hospital's multiple times, as long as we get an indeterminate form each time.

Question

1. What is $\underset{x\to \infty }{\lim }\frac{x^2}{e^x}$?

Solution

1.
$\underset{x\to \infty }{\lim }\frac{x^2}{e^x}=\frac{{\infty }^2}{e^{\infty }}=\frac{\infty }{\infty }$. Therefore, we can use L'Hospital's rule.

$\underset{x\to \infty }{\lim }\frac{x^2}{e^x}=\underset{x\to \infty }{\lim }\frac{\frac{d}{dx}(x^2)}{\frac{d}{dx}(e^x)}=\underset{x\to \infty }{\lim }\frac{2x}{e^x}=\frac{2\infty }{e^{\infty }}=\frac{\infty }{\infty }$. Therefore, we can use L'Hospital's rule again.

$\underset{x\to \infty }{\lim }\frac{2x}{e^x}=\underset{x\to \infty }{\lim }\frac{\frac{d}{dx}(2x)}{\frac{d}{dx}(e^x)}=\underset{x\to \infty }{\lim }\frac{2}{e^x}=\frac{2}{e^{\infty }}=\frac{2}{\infty }=0$

Therefore, our final answer is the limit equals $0$.

Common Mistake

L'Hospital's rule states that if:

$\underset{x\to a}{\lim }\frac{f(x)}{g(x)}=\frac{0}{0}$ or $\frac{\infty }{\infty }$ (aka an indeterminate form)

Then:

$\underset{x\to a}{\lim }\frac{f(x)}{g(x)}=\underset{x\to a}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}$.

Please note that:

$\underset{x\to a}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}$ does not equal $\underset{x\to a}{\lim }\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right)$. We are not using the quotient rule with $f(x)$ and $g(x)$!

Trigonometric Limit Shortcuts (Optional)

Some people may have learned this in Unit 1, but there are 4 trigonometric limit shortcuts that you should be aware of. When they can be used, they can save a lot of time. The 4 shortcuts are below:

1. $\underset{x\to 0}{\lim }\frac{\sin (ax)}{bx}=\frac{a}{b}$

2. $\underset{x\to 0}{\lim }\frac{ax}{\sin (bx)}=\frac{a}{b}$

3. $\underset{x\to 0}{\lim }\frac{1-\cos x}{x}=0$

4. $\underset{x\to 0}{\lim }\frac{\cos x-1}{x}=0$

Note that all of the above shortcuts can be verified with L'Hospital's rule. If time permits, I recommend verifying the shortcuts above.

The shortcuts can often allow us to skip using L'Hospital's rule. For example, let's revisit one of the problems from above.

Question

1. What is $\underset{x\to 0}{\lim }\frac{\sin (4x)}{9x}$?

Solution

We can recognize that this is in the form of $\underset{x\to 0}{\lim }\frac{\sin (ax)}{bx}$. Using the shortcut, we know this limit is equal to $\frac{a}{b}$, or in this case, $\frac{4}{9}$. 

Practice Problems