L'Hopital's rule for indeterminate forms — AP Calculus AB Study Guide

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: This chapter covers the statement of L'Hopital’s Rule, identifying 0/0 and ∞/∞ indeterminate forms, converting 0·∞ and ∞−∞ indeterminate forms, and applying repeated L'Hopital's Rule to evaluate complex limits for AP Calculus AB.

You should already know: How to evaluate one-sided and two-sided limits of functions, how to compute derivatives of all common function types including the chain rule, basic algebraic manipulation of rational and transcendental functions.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus AB style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is L'Hopital's rule for indeterminate forms?

L'Hopital's Rule is a differentiation-based technique for evaluating limits of functions that result in indeterminate forms—expressions that have no fixed well-defined value and cannot be evaluated via direct substitution. According to the AP Calculus AB Course and Exam Description (CED), this topic accounts for approximately 3-6% of total exam score, and it appears in both multiple-choice (MCQ) and free-response (FRQ) sections.

Indeterminate forms like 0/0 or ∞/∞ give no information about the actual limit of the function, while basic algebraic techniques like factoring or rationalizing only work for simple cases and fail for more complicated limits involving transcendental functions (exponentials, logarithms, trigonometric functions). L'Hopital's Rule connects the limit of a ratio of two functions to the limit of the ratio of their derivatives, which is almost always easier to evaluate. On the AP exam, you will be expected to recognize when the rule can be applied, apply it correctly, and adjust non-standard indeterminate forms to fit the rule's requirements.

2. L'Hopital's Rule for the 0/0 Indeterminate Form

The most common indeterminate form tested on the AP exam is 0/0, which occurs when both the numerator and denominator of a ratio approach 0 as $x$ approaches the limit point. The formal statement of L'Hopital's Rule for this case is:

Suppose ${\lim }_{x\to a}f(x)=0$ and ${\lim }_{x\to a}g(x)=0$, where $a$ can be a finite number, $+\infty$, or $-\infty$. If $f$ and $g$ are differentiable near $a$ (except possibly at $a$ itself), and $g^{\prime }(x)\ne 0$ near $a$, then: $$\underset{x\to a}{\lim }\frac{f(x)}{g(x)}=\underset{x\to a}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}$$ as long as the limit on the right exists (as a finite number, $+\infty$, or $-\infty$).

The intuition for this rule is straightforward: when both functions approach 0 near $a$, their behavior is well approximated by their tangent lines at $a$: $f(x)\approx f^{\prime }(a)(x-a)$ and $g(x)\approx g^{\prime }(a)(x-a)$. The $(x-a)$ terms cancel out, leaving the ratio $\frac{f^{\prime }(a)}{g^{\prime }(a)}$, which matches the result from L'Hopital's Rule. L'Hopital's Rule can be applied repeatedly if the new limit after the first application is still 0/0.

Worked Example

Evaluate ${\lim }_{x\to 0}\frac{e^{3x}-1-3x}{x^2}$.

1. First confirm the indeterminate form: substitute $x=0$: numerator $e^0-1-0=1-1=0$, denominator $0^2=0$, so 0/0 and L'Hopital's applies.
2. Differentiate numerator and denominator separately: $f^{\prime }(x)=3e^{3x}-3$, $g^{\prime }(x)=2x$.
3. Evaluate the new limit: ${\lim }_{x\to 0}\frac{3e^{3x}-3}{2x}={\lim }_{x\to 0}\frac{3(e^{3x}-1)}{2x}$. Substituting $x=0$ gives 0/0, so we apply L'Hopital's again.
4. Differentiate again: $f^{\prime \prime }(x)=9e^{3x}$, $g^{\prime \prime }(x)=2$.
5. Evaluate the new limit: ${\lim }_{x\to 0}\frac{9e^{3x}}{2}=\frac{9e^0}{2}=\frac{9}{2}$.

The original limit is $\frac{9}{2}$.

Exam tip: Always explicitly confirm you have an indeterminate 0/0 or ∞/∞ form before writing L'Hopital's Rule on an FRQ. AP readers require this justification for full points, even if you get the correct final answer.

3. L'Hopital's Rule for the ∞/∞ Indeterminate Form

The second core indeterminate form that fits directly into L'Hopital's Rule is ∞/∞, which occurs when both the numerator and denominator approach positive or negative infinity as $x$ approaches the limit point. The conditions and formula for L'Hopital's Rule are identical to the 0/0 case, only the limits of $f(x)$ and $g(x)$ change.

Formally, if ${\lim }_{x\to a}f(x)=\pm \infty$ and ${\lim }_{x\to a}g(x)=\pm \infty$, with the same differentiability and non-zero derivative conditions as before, then: $$\underset{x\to a}{\lim }\frac{f(x)}{g(x)}=\underset{x\to a}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}$$ The intuition here is that when both functions grow without bound, the rate of growth (given by the derivative) determines which function dominates, so the ratio of derivatives gives the correct limit. This form is commonly used to find limits at infinity for ratios of transcendental functions, and to identify horizontal asymptotes of complex functions. Like the 0/0 form, it can be applied repeatedly.

Worked Example

Evaluate ${\lim }_{x\to +\infty }\frac{4x^3-2x+5}{e^{4x}}$.

1. Check the form: as $x\to +\infty$, the numerator $4x^3-2x+5\to +\infty$ and the denominator $e^{4x}\to +\infty$, so we have an ∞/∞ indeterminate form, L'Hopital's applies.
2. Differentiate numerator and denominator separately: $f^{\prime }(x)=12x^2-2$, $g^{\prime }(x)=4e^{4x}$.
3. New limit: ${\lim }_{x\to +\infty }\frac{12x^2-2}{4e^{4x}}={\lim }_{x\to +\infty }\frac{3x^2-0.5}{e^{4x}}$, which is still ∞/∞. Apply L'Hopital's again.
4. Differentiate again: $f^{\prime \prime }(x)=6x$, $g^{\prime \prime }(x)=4e^{4x}$. New limit is still ∞/∞, apply L'Hopital's a third time.
5. Differentiate a third time: $f^{\prime \prime \prime }(x)=6$, $g^{\prime \prime \prime }(x)=16e^{4x}$. Evaluate the limit: ${\lim }_{x\to +\infty }\frac{6}{16e^{4x}}=0$.

The original limit is $0$.

Exam tip: For limits at infinity, remember that positive exponential functions grow faster than any polynomial, and logarithmic functions grow slower than any positive power of $x$. This lets you predict the result of repeated L'Hopital applications quickly on MCQs.

4. Converting 0·∞ and ∞−∞ Indeterminate Forms

Not all indeterminate forms are directly 0/0 or ∞/∞, but they can almost always be rewritten algebraically to fit L'Hopital's Rule. The two non-standard indeterminate forms tested on AP Calculus AB are 0·∞ (product of a function approaching 0 and a function approaching infinity) and ∞−∞ (difference of two functions both approaching infinity).

For 0·∞, rewrite the product as a ratio by moving one term to the denominator: $f(x)g(x)=\frac{f(x)}{1/g(x)}$, which becomes 0/0 if $f\to 0$ and $g\to \infty$, or $\frac{g(x)}{1/f(x)}$ which becomes ∞/∞. You should always choose the form that results in simpler differentiation. For ∞−∞, the most common conversion is to get a common denominator, which turns the difference into a single ratio that is usually 0/0.

Worked Example

Evaluate ${\lim }_{x\to 0^{+}}4x\ln x$.

1. Identify the form: as $x\to 0^{+}$, $4x\to 0$ and $\ln x\to -\infty$, so this is an indeterminate $0\cdot (-\infty )$ form that needs conversion.
2. Rewrite as a ratio: move $x$ to the denominator to get ${\lim }_{x\to 0^{+}}\frac{4\ln x}{x^{-1}}$, which is $-\infty /+\infty$, a valid ∞/∞ form for L'Hopital's Rule.
3. Differentiate numerator and denominator separately: derivative of numerator is $\frac{4}{x}$, derivative of denominator is $-x^{-2}=-\frac{1}{x^2}$.
4. Simplify the ratio: $\frac{4/x}{-1/x^2}=\frac{4}{x}\cdot (-x^2)=-4x$.
5. Evaluate the limit: ${\lim }_{x\to 0^{+}}-4x=0$.

The original limit is $0$.

Exam tip: When converting 0·∞, never put the logarithmic or exponential term in the denominator. This will always result in a much more complicated derivative that leads to unnecessary algebraic errors. Keep simpler transcendental terms in the numerator.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Differentiating the entire ratio using the quotient rule instead of differentiating numerator and denominator separately. For example, turning ${\lim }_{x\to 0}\frac{e^{2x}-1}{x}$ into the derivative of the whole ratio instead of the ratio of separate derivatives. Why: Students associate L'Hopital's Rule with differentiation, so they confuse derivative of a ratio with L'Hopital's ratio of derivatives. Correct move: Always label $f(x)$ as the numerator and $g(x)$ as the denominator, compute $f^{\prime }(x)$ and $g^{\prime }(x)$ separately, then form the new ratio before taking the limit.
• Wrong move: Applying L'Hopital's Rule to a determinate limit, for example ${\lim }_{x\to 1}\frac{x^2+1}{x+1}={\lim }_{x\to 1}\frac{2x}{1}=2$, when the correct value is $1$. Why: Students get into the habit of using L'Hopital's for all limits and skip checking the indeterminate form. Correct move: Always substitute the limit point into the numerator and denominator first, and write down the indeterminate form to confirm before applying L'Hopital's.
• Wrong move: Forgetting the chain rule when differentiating composite functions, for example differentiating $e^{3x}$ to get $e^{3x}$ instead of $3e^{3x}$. Why: Students rush differentiation after confirming the indeterminate form, and miss the inner derivative. Correct move: After writing derivatives of the numerator and denominator, pause to check every composite function for a chain rule factor before proceeding.
• Wrong move: Stopping after one application of L'Hopital's Rule when the new limit is still indeterminate, and incorrectly concluding the limit does not exist. Why: Students forget the rule can be applied repeatedly. Correct move: After each application, check the form again. If it is still 0/0 or ∞/∞, apply the rule again; only stop when you get a determinate form.
• Wrong move: Misclassifying $\infty \cdot \infty$ as indeterminate and applying L'Hopital's Rule, resulting in an incorrect sign or value. Why: Students think all combinations of infinity are indeterminate. Correct move: Remember only 0·∞ is indeterminate; ∞·∞, ∞ + ∞, and 0·0 are all determinate, so no L'Hopital's Rule is needed.

6. Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

What is the value of ${\lim }_{x\to \frac{\pi }{2}}\frac{\cos x}{\frac{\pi }{2}-x}$? A) $-1$ B) $-\frac{1}{2}$ C) $\frac{1}{2}$ D) $1$

Worked Solution: First confirm the indeterminate form: substitute $x=\frac{\pi }{2}$: numerator $\cos \left(\frac{\pi }{2}\right)=0$, denominator $\frac{\pi }{2}-\frac{\pi }{2}=0$, so we have 0/0 and L'Hopital's Rule applies. Differentiate numerator: $\frac{d}{dx}\cos x=-\sin x$. Differentiate denominator: $\frac{d}{dx}\left(\frac{\pi }{2}-x\right)=-1$. Evaluate the new limit: ${\lim }_{x\to \frac{\pi }{2}}\frac{-\sin x}{-1}=\sin \left(\frac{\pi }{2}\right)=1$. The correct answer is D.

Question 2 (Free Response)

Let $f(x)=\frac{5\ln x}{x}$ for $x>0$. (a) Evaluate ${\lim }_{x\to 1}f(x)$. Justify your answer. (b) Evaluate ${\lim }_{x\to +\infty }f(x)$. Justify your answer. (c) Find all horizontal asymptotes of $f(x)$. Justify your answer.

Worked Solution: (a) Substitute $x=1$: numerator $5\ln (1)=0$, denominator $1$, so this is the determinate form $\frac{0}{1}=0$. L'Hopital's Rule does not apply. ${\lim }_{x\to 1}f(x)=0$. (b) As $x\to +\infty$, $5\ln x\to +\infty$ and $x\to +\infty$, so we have an ∞/∞ indeterminate form that meets the conditions for L'Hopital's Rule. Differentiate numerator: $\frac{d}{dx}(5\ln x)=\frac{5}{x}$. Differentiate denominator: $\frac{d}{dx}(x)=1$. The new limit is ${\lim }_{x\to +\infty }\frac{5}{x}=0$, so ${\lim }_{x\to +\infty }f(x)=0$. (c) A horizontal asymptote exists at $y=L$ if ${\lim }_{x\to +\infty }f(x)=L$ or ${\lim }_{x\to -\infty }f(x)=L$. Since $f(x)$ is only defined for $x>0$, we only need the limit at positive infinity, which we found to be $0$. So the only horizontal asymptote is $y=0$.

Question 3 (Application / Real-World Style)

The number of social media followers of a new brand $t$ months after launch is modeled by $F(t)=\frac{10000t}{1+0.2e^{0.5t}}$, where $F(t)$ is measured in followers. What is the limit of the follower count as $t\to +\infty$? Interpret your result in the context of the problem.

Worked Solution: Check the form as $t\to +\infty$: numerator $10000t\to +\infty$, denominator $1+0.2e^{0.5t}\to +\infty$, so we have an ∞/∞ indeterminate form, L'Hopital's Rule applies. Differentiate numerator: $\frac{d}{dt}(10000t)=10000$. Differentiate denominator: $\frac{d}{dt}(1+0.2e^{0.5t})=0.1e^{0.5t}$. The new limit is ${\lim }_{t\to +\infty }\frac{10000}{0.1e^{0.5t}}={\lim }_{t\to +\infty }\frac{100000}{e^{0.5t}}=0$. In context, this means that after the initial growth spurt, the brand's follower growth slows to near zero as the platform's audience becomes saturated, so the total number of new followers added each month approaches zero over time.

7. Quick Reference Cheatsheet

8. What's Next

L'Hopital's Rule is a foundational tool for analyzing the end behavior of functions and evaluating complex limits, which is required for all remaining topics in Unit 4 Contextual Applications of Differentiation, and later for improper integration in Unit 8. Without correctly applying L'Hopital's Rule to evaluate indeterminate limits, you will not be able to find horizontal asymptotes for transcendental functions or evaluate improper integrals, both of which are common, high-weight topics on the AP exam. L'Hopital's Rule also reinforces the core connection between derivatives and limits that forms the foundation of all differential calculus, and it simplifies evaluating limits for complex real-world models that cannot be solved with basic algebra alone. The follow-on topics you should study next are:

• Related rates problem solving
• Asymptotes and continuity
• Improper integral evaluation