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

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: The formal statement of L'Hopital's rule, identification of all 7 indeterminate forms, conversion of non-0/0 indeterminate forms to 0/0 or ∞/∞, and step-by-step application to evaluate all indeterminate limit types tested on the AP exam.

You should already know: How to evaluate one-sided and two-sided limits algebraically; how to compute derivatives of all function types including transcendental functions; how to apply the chain rule.

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 BC 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 to evaluate limits of indeterminate forms, where direct substitution of the limit value results in an expression with no defined limiting value. In the AP Calculus BC Course and Exam Description (CED), this topic falls under Unit 4: Contextual Applications of Differentiation, and accounts for approximately 2-3% of the total exam score, appearing regularly in both multiple-choice (MCQ) and free-response (FRQ) sections.

A formal statement of the rule: If ${\lim }_{x\to a}f(x)=0$ and ${\lim }_{x\to a}g(x)=0$, or ${\lim }_{x\to a}f(x)=\pm \infty$ and ${\lim }_{x\to a}g(x)=\pm \infty$, and $f$ and $g$ are differentiable on an open interval containing $a$ (except possibly at $a$ itself), and $g^{\prime }(x)\ne 0$ on that interval except possibly at $a$, then ${\lim }_{x\to a}\frac{f(x)}{g(x)}={\lim }_{x\to a}\frac{f^{\prime }(x)}{g^{\prime }(x)}$. This holds for one-sided limits ($x\to a^{+},x\to a^{-}$) and limits at infinity ($x\to \pm \infty$) as well. Indeterminate forms are expressions where competing trends (one term pulling toward 0, another toward ∞, for example) mean the original limit cannot be determined from the individual limits of the parts, and they can resolve to any finite or infinite value.

2. Indeterminate Forms 0/0 and ∞/∞

These two indeterminate forms are the only forms to which L'Hopital's rule applies directly; all other indeterminate forms must be rewritten to fit one of these two structures before the rule can be used. The intuition for the rule comes from linear approximation of differentiable functions near the limit point. If both $f(a)=0$ and $g(a)=0$, near $x=a$, $f(x)\approx f^{\prime }(a)(x-a)$ and $g(x)\approx g^{\prime }(a)(x-a)$, so their ratio simplifies to $f^{\prime }(a)/g^{\prime }(a)$, which is exactly what L'Hopital's rule gives. For limits at infinity, the same intuition holds: derivatives capture how each function grows as $x$ becomes very large, so their relative growth rates are given by the ratio of derivatives.

A key property of the rule is that it can be applied repeatedly: if after taking the first derivatives you still get an indeterminate 0/0 or ∞/∞, you can apply it again, as long as the conditions hold each time. The only requirement is that every application confirms the new limit is still indeterminate.

Worked Example

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

1. Check indeterminacy by direct substitution: $e^0-1-2(0)=0$ for the numerator, denominator is $0^2=0$, so we have valid 0/0 indeterminate form, can apply L'Hopital's rule.
2. Differentiate numerator and denominator separately: $f^{\prime }(x)=2e^{2x}-2$, $g^{\prime }(x)=2x$.
3. Evaluate the new limit: ${\lim }_{x\to 0}\frac{2e^{2x}-2}{2x}=\frac{2(1)-2}{0}=0/0$, which is still indeterminate, so apply L'Hopital's rule again.
4. Differentiate again: $f^{\prime \prime }(x)=4e^{2x}$, $g^{\prime \prime }(x)=2$.
5. Evaluate the new limit: ${\lim }_{x\to 0}\frac{4e^{2x}}{2}=\frac{4(1)}{2}=2$, so the original limit equals 2.

Exam tip: Always confirm the indeterminate form explicitly on FRQ answers; AP graders require this step to award full points for using L'Hopital's rule.

3. Indeterminate Products $0\cdot \infty$

The next most common indeterminate form is a product where one term approaches 0 and the other approaches ±∞, written $0\cdot \infty$. This is indeterminate because 0 pulls the product toward 0 while ∞ pulls it toward ∞, so the result can be any finite value, 0, or ∞. To apply L'Hopital's rule, we rewrite the product as a fraction by moving one term to the denominator: if we have ${\lim }_{x\to a}f(x)g(x)$ where $\lim f=0$ and $\lim g=\infty$, we can rewrite this as either ${\lim }_{x\to a}\frac{f(x)}{1/g(x)}$ (which becomes 0/0) or ${\lim }_{x\to a}\frac{g(x)}{1/f(x)}$ (which becomes ∞/∞). Both conversions are mathematically valid, but one is almost always simpler to differentiate than the other, so you should always choose the conversion that minimizes extra work like quotient rule.

Worked Example

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

1. Check the form: as $x\to 0^{+}$, $x\to 0$ and $\ln (4x)\to -\infty$, so we have an indeterminate product $0\cdot (-\infty )$.
2. Rewrite as a fraction: move $x$ to the denominator to get ${\lim }_{x\to 0^{+}}\frac{\ln (4x)}{1/x}$, which becomes $-\infty /\infty$, a valid form for L'Hopital's. (Moving $\ln (4x)$ to the denominator would give $\frac{x}{1/\ln (4x)}$, which requires differentiating a reciprocal logarithm, a much messier process, so this conversion is preferred.)
3. Differentiate numerator and denominator separately: $\frac{d}{dx}[\ln (4x)]=\frac{1}{x}$, $\frac{d}{dx}[1/x]=-\frac{1}{x^2}$.
4. Simplify the ratio of derivatives: $\frac{1/x}{-1/x^2}=-x$.
5. Evaluate the limit: ${\lim }_{x\to 0^{+}}(-x)=0$, so the original limit is 0.

Exam tip: When converting a product, always leave the simpler term to differentiate in the numerator to avoid introducing extra chain rule or quotient rule errors.

4. Indeterminate Differences $\infty -\infty$

Indeterminate differences occur when we have the difference of two terms, both approaching +∞ or both approaching -∞, written $\infty -\infty$. This is indeterminate because the two infinite terms compete, and the result can be 0, any finite number, or ±∞. To solve this type of limit, we need to convert the difference into a single fraction, almost always by combining terms over a common denominator, factoring, or multiplying by a conjugate to eliminate radicals. Once combined, the resulting fraction will almost always be a 0/0 or ∞/∞ indeterminate form that L'Hopital's rule can be applied to. This form appears commonly in limits of rational functions with different denominators and limits involving logarithmic expressions.

Worked Example

Evaluate ${\lim }_{x\to 1}\left(\frac{1}{x-1}-\frac{1}{\ln x}\right)$.

1. Check the form: as $x\to 1$, both $\frac{1}{x-1}$ and $\frac{1}{\ln x}$ approach ±∞, so this is an indeterminate difference $\infty -\infty$.
2. Combine the fractions over a common denominator to get a single fraction: ${\lim }_{x\to 1}\frac{\ln x-(x-1)}{(x-1)\ln x}$.
3. Check indeterminacy: numerator at $x=1$ is $\ln 1-0=0$, denominator is $(0)(\ln 1)=0$, so we have 0/0, valid for L'Hopital's.
4. Differentiate numerator and denominator: numerator derivative is $\frac{1}{x}-1=\frac{1-x}{x}$, denominator derivative (via product rule) is $\ln x+(x-1)\cdot \frac{1}{x}=\ln x+1-\frac{1}{x}$.
5. Substitute $x=1$: $\frac{0}{0+1-1}=\frac{0}{0}$, still indeterminate, so apply L'Hopital's again.
6. Differentiate again: numerator derivative is $-\frac{1}{x^2}$, denominator derivative is $\frac{1}{x}+\frac{1}{x^2}$.
7. Evaluate the limit: ${\lim }_{x\to 1}\frac{-1/x^2}{1/x+1/x^2}=\frac{-1}{1+1}=-\frac{1}{2}$, so the original limit is $-\frac{1}{2}$.

Exam tip: Never differentiate each term of the difference separately; you must always combine into a single fraction first before applying L'Hopital's rule.

5. Indeterminate Powers $0^0,1^{\infty },{\infty }^0$

These three indeterminate forms are exponential expressions of the form ${\lim }_{x\to a}f(x{)}^{g(x)}$, where the limits of the base and exponent create one of these three indeterminate combinations. For example, $0^0$ is indeterminate because the base pulls the expression toward 0 while the exponent pulls it toward $e^0=1$, so the result depends on the relative speed of the two trends. To solve these, we use the natural logarithm to convert the exponential into a product, which we can then convert to 0/0 or ∞/∞ to apply L'Hopital's. The standard step-by-step process is: (1) Let $L=\lim f(x{)}^{g(x)}$, (2) Take natural log of both sides: $\ln L=\lim g(x)\ln f(x)$, (3) Solve the resulting indeterminate product limit, (4) Exponentiate to get $L=e^{\ln L}$.

Worked Example

Evaluate ${\lim }_{x\to 0^{+}}(3x{)}^x$.

1. Check the form: as $x\to 0^{+}$, $3x\to 0$ and $x\to 0$, so we have $0^0$, an indeterminate power.
2. Let $L={\lim }_{x\to 0^{+}}(3x{)}^x$, so $\ln L={\lim }_{x\to 0^{+}}x\ln (3x)$.
3. This is an indeterminate product $0\cdot (-\infty )$. Rewrite as ${\lim }_{x\to 0^{+}}\frac{\ln (3x)}{1/x}$, which is $-\infty /\infty$, valid for L'Hopital's.
4. Differentiate numerator and denominator: numerator derivative is $\frac{1}{x}$, denominator derivative is $-\frac{1}{x^2}$. Simplify the ratio: $\frac{1/x}{-1/x^2}=-x$.
5. Evaluate the limit: $\ln L={\lim }_{x\to 0^{+}}(-x)=0$.
6. Exponentiate to get the original limit: $L=e^0=1$.

Exam tip: Don't forget to undo the natural logarithm at the end; forgetting the final exponentiation is one of the most common mistakes AP graders see on this question type.

6. Common Pitfalls (and how to avoid them)

• Wrong move: Applying L'Hopital's rule to a determinate form, e.g., evaluating ${\lim }_{x\to 0}\frac{x}{x+1}$ by differentiating to get 1/1 = 1, when direct substitution gives 0/1 = 0. Why: Students get in the habit of using L'Hopital's for every limit and forget to check the indeterminacy condition first. Correct move: Always plug in the limit value first to confirm you have an indeterminate form before applying L'Hopital's.
• Wrong move: Differentiating the entire fraction using the quotient rule, instead of differentiating numerator and denominator separately, e.g., differentiating ${\lim }_{x\to 0}\frac{e^x-1}{x}$ as $\frac{x{e}^x-(e^x-1)}{x^2}$. Why: Confusion between L'Hopital's rule and the derivative quotient rule, since we work with a ratio of functions. Correct move: Explicitly label $f(x)$ (numerator) and $g(x)$ (denominator) on scratch paper before differentiating to avoid mixing up rules.
• Wrong move: Stopping after one application of L'Hopital's when the result is still indeterminate, leaving the answer as 0/0. Why: Students assume one differentiation is enough and don't check the new limit for indeterminacy. Correct move: After each differentiation step, substitute the limit value to check for indeterminacy; apply L'Hopital's again if the result is still indeterminate.
• Wrong move: Forgetting to exponentiate after using the logarithm for an indeterminate power, leaving the answer as $\ln L=2$ instead of $L=e^2$. Why: Students get focused on applying L'Hopital's to the product after taking the log and forget the original limit is for the power, not the log of the power. Correct move: Write "Original limit $L=e^{\ln L}$" explicitly before solving for $\ln L$ to remind yourself of the final step.
• Wrong move: Applying L'Hopital's rule directly to discrete sequences, e.g., evaluating ${\lim }_{n\to \infty }\frac{\ln n}{n}$ on the discrete sequence without extending to a continuous function. Why: The rule only applies to differentiable functions, which sequences are not. Correct move: When evaluating a sequence limit, restate it as the limit of the corresponding continuous function as $x\to \infty$, apply L'Hopital's to the continuous version, then conclude the sequence limit matches.

7. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

What is the value of ${\lim }_{x\to \infty }\frac{5x^3+2x-1}{3x^3-4x^2}$? A) $0$ B) $\frac{5}{3}$ C) $\infty$ D) $\frac{3}{5}$

Worked Solution: First confirm the form: as $x\to \infty$, the numerator approaches $\infty$ and the denominator approaches $\infty$, so we have an indeterminate $\infty /\infty$ form, valid for L'Hopital's rule. Differentiate numerator and denominator to get $\frac{15x^2+2}{9x^2-8x}$, which is still $\infty /\infty$. Apply L'Hopital's again to get $\frac{30x}{18x-8}$, which is still $\infty /\infty$. Apply L'Hopital's a third time to get $\frac{30}{18}=\frac{5}{3}$, which is determinate. The correct answer is B.

Question 2 (Free Response)

Consider the function $f(x)={\left(1+\frac{k}{x}\right)}^x$ for $x>0$ and constant $k>0$. (a) Show that ${\lim }_{x\to \infty }f(x)$ is an indeterminate form of type $1^{\infty }$, then evaluate the limit. (b) Use your result from part (a) to find ${\lim }_{x\to \infty }{\left(1+\frac{2}{x}\right)}^{3x}$. (c) For what value of $k$ is ${\lim }_{x\to \infty }{\left(1+\frac{k}{x}\right)}^x=e^5$?

Worked Solution: (a) As $x\to \infty$, $1+\frac{k}{x}\to 1$ and $x\to \infty$, so the form is $1^{\infty }$, which is indeterminate. Let $L={\lim }_{x\to \infty }{\left(1+\frac{k}{x}\right)}^x$, so $\ln L={\lim }_{x\to \infty }x\ln \left(1+\frac{k}{x}\right)={\lim }_{x\to \infty }\frac{\ln \left(1+\frac{k}{x}\right)}{1/x}$. This is 0/0, so apply L'Hopital's: numerator derivative is $\frac{1}{1+k/x}\cdot \left(-\frac{k}{x^2}\right)$, denominator derivative is $-1/x^2$. Cancel terms to get ${\lim }_{x\to \infty }\frac{k}{1+k/x}=k$, so $\ln L=k$ and $L=e^k$. (b) Rewrite the limit as ${\left({\lim }_{x\to \infty }{\left(1+\frac{2}{x}\right)}^x\right)}^3=(e^2{)}^3=e^6$. (c) From part (a), $e^k=e^5$, so $k=5$.

Question 3 (Application / Real-World Style)

In population genetics, the expected time to fixation of a new neutral allele starting from an initial frequency $p$ in a population of constant size $N$ is given by $T(p)=\frac{-2N}{p}(1-p)\ln (1-p)$. What is the limit of $T(p)$ as the initial frequency $p$ approaches 0 from the positive side? Give your answer in terms of $N$ and interpret the result.

Worked Solution: We evaluate ${\lim }_{p\to 0^{+}}\frac{-2N(1-p)\ln (1-p)}{p}$. As $p\to 0^{+}$, numerator $\to 0$ and denominator $\to 0$, so 0/0 indeterminate. Factor out the constant $-2N$ to get $-2N{\lim }_{p\to 0^{+}}\frac{(1-p)\ln (1-p)}{p}$. Apply L'Hopital's: numerator derivative is $-\ln (1-p)-1$, denominator derivative is 1. Evaluate the limit: $-\ln (1)-1=-1$. Multiply by the constant: $-2N(-1)=2N$. Interpretation: When a new neutral allele is very rare in a population (initial frequency near 0), the expected time for the allele to spread to the entire population is approximately twice the constant population size $N$.

8. Quick Reference Cheatsheet

9. What's Next

L'Hopital's rule is a foundational tool for evaluating indeterminate limits, required for multiple upcoming topics in AP Calculus BC. Immediately after this topic, you will use L'Hopital's to evaluate limits for improper integrals in Unit 8, where most improper integrals require evaluating indeterminate limits at infinity or near discontinuities. Beyond Unit 4, L'Hopital's is used to compare growth rates of different functions, a key skill for analyzing convergence of infinite series in Unit 10. It also appears frequently in contextual problems that require analyzing the limiting behavior of rates of change. Without mastering the ability to correctly identify indeterminate forms and apply L'Hopital's rule, you will lose points on multiple questions across later units of the course.

Improper Integrals Infinite Series Convergence Tests Growth Rate Comparison of Functions