AP Calculus AB and BC

L'Hopital's Rule: When You Can and Cannot Use It

Q: Can you use L'Hopital's Rule for 0 times infinity?

Not directly. The forms 0 \cdot \infty and \infty - \infty are excluded from both the AB and BC exams. In other coursework you would first rewrite the expression as a single fraction so that a \frac{0}{0} or \frac{\infty}{\infty} form appears, then apply the rule.

Q: How many times can you apply L'Hopital's Rule?

As many times as the quotient stays \frac{0}{0} or \frac{\infty}{\infty}. Re-check the form after each pass and stop the instant it resolves to a determinate value.

L'Hopital's Rule applies only when direct substitution gives 0/0 or infinity/infinity. Verify the form first: a value like 1/1, or a 5/0 form whose limit is infinite or does not exist, means the rule does not apply and gives a wrong answer. When it applies, differentiate top and bottom separately, then re-evaluate.

Verify the Form Before You Touch the Rule

L'Hopital's Rule is a shortcut for evaluating a limit of a quotient $\frac{f(x)}{g(x)}$ , but it comes with one hard precondition: direct substitution has to produce the indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ . Checking that form is not a formality you can skip when you are in a hurry. It is the whole rule of admission. If substitution gives anything else, the rule is not valid, and applying it can convert a correct limit into a wrong number. This is Topic 4.7 on the AP Calculus AB and BC exams.

The reason the check matters is that the rule is a theorem with a hypothesis, not a universal identity. It says that when $\frac{f(x)}{g(x)}$ approaches $\frac{0}{0}$ or $\frac{\infty}{\infty}$ , the original limit equals $\lim_{x \to a} \frac{f'(x)}{g'(x)}$ , provided that second limit exists or is infinite. Strip away the hypothesis and the two limits have no reason to be equal. That is why an unchecked application does not merely risk an error: it computes the limit of a genuinely different expression.

0/0 is a label, not a value

The College Board is explicit that $\frac{0}{0}$ names an indeterminate form; it is not a fraction you can simplify to $1$ or to $0$ . Treating it as a number is the exact error Topic 4.7 warns against. It signals only that more work is needed, not that the answer is decided.

How to Recognize an Indeterminate Form

Recognition is mechanical. Substitute the value the variable approaches into the numerator and the denominator on their own, then read the pair of results. Two outcomes, and only two, let you use L'Hopital's Rule: the numerator and denominator both go to $0$ , or they both grow without bound. Every other outcome already tells you the limit directly, so there is nothing left to resolve and no reason to differentiate.

Substitution result	Indeterminate?	L'Hopital's allowed?
$\frac{0}{0}$	Yes	Yes
$\frac{\infty}{\infty}$	Yes	Yes
$\frac{5}{0}$	No (limit is $\pm\infty$ or does not exist)	No
$\frac{0}{5}$	No (limit is $0$ )	No
$\frac{3}{2}$	No (limit is $\frac{3}{2}$ )	No

The bottom three rows are worth staring at, because they are where students lose points. A result like $\frac{5}{0}$ means the quotient blows up: the limit is infinite or does not exist, and that is the final answer. A result like $\frac{0}{5}$ equals $0$ , and $\frac{3}{2}$ equals $\frac{3}{2}$ . In each case substitution handed you the answer. Differentiating the top and bottom would replace that answer with the limit of a different function entirely.

A couple of forms tempt students into L'Hopital's Rule even though they are not quotients yet: $0 \cdot \infty$ and $\infty - \infty$ . The AP Course and Exam Description (CED) explicitly excludes these from both the AB and BC exams, so you will not be asked to resolve them. If you ever meet one in other coursework, you first rewrite it as a single fraction; only then can a $\frac{0}{0}$ or $\frac{\infty}{\infty}$ form appear.

Applying L'Hopital's Rule

Once the form is confirmed, the mechanics are short. Differentiate the numerator and the denominator separately, keep them as a quotient, and evaluate again. The single most common execution error is reaching for the quotient rule. L'Hopital's Rule does not differentiate $\frac{f(x)}{g(x)}$ as one object; it differentiates $f$ and $g$ independently and then divides the results.

\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \quad \text{when } \frac{f(x)}{g(x)} \to \frac{0}{0} \text{ or } \frac{\infty}{\infty}

After you form the new quotient $\frac{f'(x)}{g'(x)}$ , substitute again. If you now get a determinate value, that value is the limit and you are done. If you get $\frac{0}{0}$ or $\frac{\infty}{\infty}$ a second time, the rule still applies, so verify the form and differentiate again. You can repeat this as many times as the form stays indeterminate, but you must re-check after every pass, because the moment the form resolves you must stop.

One caution on repetition: L'Hopital's Rule promises the answer only when the new limit $\lim_{x \to a} \frac{f'(x)}{g'(x)}$ actually exists or runs to $\pm\infty$ . On the AB and BC exams the problems are built so that a few passes resolve cleanly, so if your quotient keeps returning an indeterminate form and never simplifies, that is a signal to try algebra first, such as factoring or dividing by the dominant term, rather than differentiating forever.

Common Misuse: When You Cannot Use It

Almost every wrong use of L'Hopital's Rule traces back to skipping the form check. Here are the misuses that show up most on quizzes and free-response scoring:

The form is not indeterminate. If substitution gives a defined value, a nonzero-over-zero like $\frac{4}{0}$ , or a zero-over-nonzero like $\frac{0}{7}$ , the limit is already settled. Differentiating changes the problem and usually produces a wrong answer, as the last worked example shows.
Using the quotient rule by mistake. Differentiate the numerator and denominator on their own. The derivative of the whole quotient is a different expression and gives the wrong limit.
Forgetting to re-verify between passes. Each application is legal only if the current quotient is still $\frac{0}{0}$ or $\frac{\infty}{\infty}$ . Applying it once more after the form has resolved reintroduces the same error.
Reaching for it when simpler tools work. A limit like $\frac{3x^2 + 2x}{x^2 - 5}$ as $x \to \infty$ is faster to settle by dividing by the highest power, and rational limits at a finite point often factor and cancel. L'Hopital's Rule is legal on many of these, but it is not always the quickest or safest route.

The habit that keeps you safe is boring and reliable: before you differentiate anything, substitute and name the form out loud. If it is $\frac{0}{0}$ or $\frac{\infty}{\infty}$ , proceed. If it is anything else, you already have your answer or you need a different technique. That one check is what separates a limit L'Hopital's Rule can finish from one it will quietly ruin, and it costs you only a few seconds under exam pressure. The verification step is the method.

Worked examples

Worked example

A 0/0 form

Evaluate $\lim_{x \to 0} \frac{e^x - 1}{x}$ .

Check the form by substituting $x = 0$ : $\frac{e^{0} - 1}{0} = \frac{1 - 1}{0} = \frac{0}{0}$ . The form is indeterminate, so the rule is allowed.
Name the pieces $f(x) = e^x - 1$ and $g(x) = x$ , then differentiate each separately: $f'(x) = e^x$ and $g'(x) = 1$ .
Replace the original quotient with $\frac{f'(x)}{g'(x)}$ and evaluate: $\lim_{x \to 0} \frac{e^x}{1} = \frac{e^{0}}{1} = \frac{1}{1} = 1$ .

Worked example

An infinity/infinity form needing two passes

Evaluate $\lim_{x \to \infty} \frac{x^2}{e^x}$ .

Read off the behavior as $x \to \infty$ : the numerator $x^2 \to \infty$ and the denominator $e^x \to \infty$ , giving $\frac{\infty}{\infty}$ . This is indeterminate, so apply the rule.
With $f(x) = x^2$ and $g(x) = e^x$ , differentiate separately: $f'(x) = 2x$ and $g'(x) = e^x$ . The new limit is $\lim_{x \to \infty} \frac{2x}{e^x}$ .
Re-check the form: as $x \to \infty$ , $\frac{2x}{e^x} \to \frac{\infty}{\infty}$ , still indeterminate. Apply the rule a second time. Differentiating $2x$ gives $2$ and differentiating $e^x$ gives $e^x$ , so the limit becomes $\lim_{x \to \infty} \frac{2}{e^x}$ .
Evaluate the last form: as $x \to \infty$ , $e^x \to \infty$ , so $\frac{2}{e^x} \to \frac{2}{\infty} = 0$ . This is a determinate value, so stop.

Worked example

A limit that looks like a job for L'Hopital's but is not

Evaluate $\lim_{x \to 0} \frac{\cos x}{x + 1}$ , and see why L'Hopital's Rule gives the wrong answer here.

Always check the form first. Substitute $x = 0$ : $\frac{\cos 0}{0 + 1} = \frac{1}{1} = 1$ . The form is $\frac{1}{1}$ , a defined value, so the limit is simply $1$ . Direct substitution already finished the problem.
Because the form is not $\frac{0}{0}$ or $\frac{\infty}{\infty}$ , L'Hopital's Rule does not apply. Watch what happens if you use it anyway: differentiating the numerator gives $-\sin x$ and the denominator gives $1$ , producing $\lim_{x \to 0} \frac{-\sin x}{1} = \frac{-\sin 0}{1} = 0$ .
The illegal application returned $0$ , but the true limit is $1$ . The rule changed the answer because the hypothesis, an indeterminate form, was never satisfied. This is exactly the misuse to guard against.

1 by direct substitution. L'Hopital's Rule does not apply; misusing it wrongly gives 0.

Frequently asked questions

Is L'Hopital's Rule on the AP Calculus exam?

Yes. It is Topic 4.7 and appears on both AB and BC. The exam only asks you to use it on the indeterminate forms $\frac{0}{0}$ and $\frac{\infty}{\infty}$ .

Can you use L'Hopital's Rule for 0 times infinity?

Not directly. The forms $0 \cdot \infty$ and $\infty - \infty$ are excluded from both the AB and BC exams. In other coursework you would first rewrite the expression as a single fraction so that a $\frac{0}{0}$ or $\frac{\infty}{\infty}$ form appears, then apply the rule.

Do you use the quotient rule with L'Hopital's Rule?

No. Differentiate the numerator and the denominator separately, then divide. Using the quotient rule on the whole fraction produces a different expression and the wrong limit.

How many times can you apply L'Hopital's Rule?

As many times as the quotient stays $\frac{0}{0}$ or $\frac{\infty}{\infty}$ . Re-check the form after each pass and stop the instant it resolves to a determinate value.