4.7 Using L'Hopitals Rule for Determining Limits in Indeterminate Forms

L'Hospital's Rule lets you evaluate a limit that gives the indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ by taking the derivative of the top and the derivative of the bottom separately, then evaluating the new limit. Before you apply it, you have to confirm the limit actually has one of those two forms.

Why This Matters for the AP Calculus Exam

This is the closing topic of Unit 4, and it ties your derivative skills back to limits. On the AP Calculus exam you will see indeterminate-form limits in both multiple-choice and free-response settings. When a limit problem appears on a free-response question, you are expected to verify that the rule applies before using it, so showing that both the numerator and denominator approach 0 (or both approach infinity) is important for clear exam work. The skill being tested is recognizing the form, choosing the right procedure, and carrying out the derivative correctly.

Only $\frac{0}{0}$ and $\frac{\infty}{\infty}$ are assessed on the AP Calculus AB and BC exams. Other indeterminate forms like $\infty - \infty$ will not be tested.

Key Takeaways

• L'Hospital's Rule applies only when $\lim_{x\to a}\frac{f(x)}{g(x)}$ gives $\frac{0}{0}$ or $\frac{\infty}{\infty}$.
• Always check and state that the numerator and denominator each go to 0 (or each go to $\pm\infty$) before applying the rule.
• The rule says $\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}$. You take the derivative of the top and bottom separately, not the quotient rule.
• If the new limit is still indeterminate, you can apply the rule again.
• $\frac{0}{0}$ and $\frac{\infty}{\infty}$ are labels for indeterminate forms, not actual values, so do not write the limit as equal to $\frac{0}{0}$.
• The rule works for finite limits ($x\to a$) and for limits at infinity ($x\to\infty$).

L'Hospital's Rule

L'Hospital's Rule states that if $\lim_{x\to a}\frac{f(x)}{g(x)}$ gives $\frac{0}{0}$ or $\pm\frac{\infty}{\infty}$, then

$\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}$

In plain terms, you can evaluate the limit of an indeterminate form by replacing the numerator and denominator with their own derivatives.

One thing to keep in mind: $\frac{0}{0}$ and $\frac{\infty}{\infty}$ are just names for indeterminate forms. They are not numbers. So it is incorrect to write $\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{0}{0}$. Use that form only as a label for what the limit is doing, then move on to the rule.

Worked Example

Evaluate the following limit.

$\lim_{x \to \frac{\pi}{2}}\frac{\cos(x)}{x-\frac{\pi}{2}}$

Plugging $x=\frac{\pi}{2}$ into $\frac{\cos(x)}{x-\frac{\pi}{2}}$ gives the indeterminate form $\frac{0}{0}$. That signals you should consider L'Hospital's Rule.

On a free-response question, show that the numerator $f(x)$ and the denominator $g(x)$ each go to the needed value separately. Do that first:

$\lim_{x \to \frac{\pi}{2}}{\cos(x)}=0$

$\lim_{x \to \frac{\pi}{2}}\left(x-\frac{\pi}{2}\right)=0$

Since both limits equal 0, L'Hospital's Rule can be applied. Write that statement out before you actually use the rule.

Now take the derivatives of the top and bottom:

$\lim_{x \to \frac{\pi}{2}}\frac{\cos(x)}{x-\frac{\pi}{2}}=\lim_{x\to \frac{\pi}{2}}\frac{\frac{d}{dx}[\cos(x)]}{\frac{d}{dx}[x-\frac{\pi}{2}]}$

$=\lim_{x\to \frac{\pi}{2}}\frac{-\sin(x)}{1}$

$=\frac{-\sin\left(\frac{\pi}{2}\right)}{1}$

$=-1$

So the final answer is:

$\lim_{x \to \frac{\pi}{2}}\frac{\cos(x)}{x-\frac{\pi}{2}}=-1$

How to Use This on the AP Calculus Exam

Problem Solving

1. Plug in the value the variable approaches.
2. If you get $\frac{0}{0}$ or $\frac{\infty}{\infty}$, the limit is indeterminate and the rule is a candidate.
3. State that the numerator and denominator each approach 0 (or each approach $\pm\infty$).
4. Differentiate the top and bottom separately.
5. Take the limit of the new fraction. If it is still indeterminate, apply the rule again.

Free Response

When a limit shows up on a free-response question, write out the verification step. Showing $\lim f(x)=\lim g(x)=0$ (or both going to infinity) before applying the rule is important for clear, complete exam work. Skipping that step leaves your reasoning incomplete even if the final number is right.

Common Trap

Make sure the form is truly indeterminate before differentiating. If a limit already evaluates to something like $\frac{3}{0}$ or $\frac{5}{2}$, L'Hospital's Rule does not apply, and using it will give you a wrong answer.

Practice Problems

Evaluate the following limits. Treat these like free-response questions and check conditions.

Question 1

$\lim_{x\to 0}\frac{\tan(x)}{7x+\tan(x)}$

Question 2

$\lim_{x\to \infty}\frac{3x^2-8}{7x^2+21}$

Solutions

Question 1

Plugging $x=0$ into $\frac{\tan(x)}{7x+\tan(x)}$ gives the indeterminate form $\frac{0}{0}$. Since the expression mixes function types, there is no clean algebraic way to simplify it, so L'Hospital's Rule is the move.

First, show each piece goes to 0:

$\lim_{x\to 0}\tan(x)=0$

$\lim_{x\to 0}\left(7x+\tan(x)\right)=0$

Since both limits equal 0, L'Hospital's Rule can be applied.

$\lim_{x\to 0}\frac{\tan(x)}{7x+\tan(x)}=\lim_{x\to 0}\frac{\frac{d}{dx}[\tan(x)]}{\frac{d}{dx}[7x+\tan(x)]}$

$=\lim_{x\to 0}\frac{\sec^2(x)}{7+\sec^2(x)}$

$=\frac{\sec^2(0)}{7+\sec^2(0)}$

$=\frac{1}{7+1}=\frac{1}{8}$

So:

$\lim_{x\to 0}\frac{\tan(x)}{7x+\tan(x)}=\frac{1}{8}$

Question 2

Plugging in $x\to\infty$ for $\frac{3x^2-8}{7x^2+21}$ gives the indeterminate form $\frac{\infty}{\infty}$, so L'Hospital's Rule applies.

$\lim_{x\to \infty}\left(3x^2-8\right)=\infty$

$\lim_{x\to \infty}\left(7x^2+21\right)=\infty$

Since both go to infinity, L'Hospital's Rule can be applied.

$\lim_{x\to \infty}\frac{3x^2-8}{7x^2+21}=\lim_{x\to \infty}\frac{\frac{d}{dx}[3x^2-8]}{\frac{d}{dx}[7x^2+21]}$

$=\lim_{x\to \infty}\frac{6x}{14x}$

$=\frac{3}{7}$

Therefore:

$\lim_{x\to \infty}\frac{3x^2-8}{7x^2+21}=\frac{3}{7}$

Notice that in Question 2 the first application still left a ratio, but $\frac{6x}{14x}$ simplifies to $\frac{3}{7}$ directly. If a single application does not resolve the form, you are allowed to apply the rule again.

Common Misconceptions

• Using the quotient rule instead. L'Hospital's Rule takes the derivative of the numerator and the derivative of the denominator separately. It is not the derivative of the whole fraction.
• Writing the limit equal to $\frac{0}{0}$. Those symbols are labels for indeterminate forms, not values. Do not write $\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{0}{0}$.
• Skipping the verification step. You must confirm the limit is $\frac{0}{0}$ or $\frac{\infty}{\infty}$ first. Applying the rule to a non-indeterminate limit gives a wrong answer.
• Assuming the limit of a quotient equals the quotient of the limits. When the denominator's limit is 0, $\lim_{x\to a}\frac{f(x)}{g(x)}\neq\frac{\lim_{x\to a}f(x)}{\lim_{x\to a}g(x)}$, which is exactly why you need a different tool here.
• Stopping too early. If the new fraction is still indeterminate after one application, apply the rule again rather than guessing.
• Forgetting other forms are off-limits. On the AP Calculus exam, only $\frac{0}{0}$ and $\frac{\infty}{\infty}$ are tested. Forms like $\infty-\infty$ are not assessed.

Related AP Calculus Guides

• Unit 4 Overview: Contextual Applications of Differentiation
• 4.1 Interpreting the Meaning of the Derivative in Context
• 4.2 Straight-Line Motion: Connecting Position, Velocity, and Acceleration
• 4.4 Intro to Related Rates
• 4.6 Approximating Values of a Function Using Local Linearity and Linearization
• 4.5 Solving Related Rates Problems