# L'Hopital's Rule for Indeterminate Forms

> AP Calculus BC · Unit 4: Contextual Applications of Differentiation
> Source: https://www.owlsprep.com/study/ap-calculus-bc-u4-l-hopital-s-rule-for/

This module covers L'Hopital's rule for indeterminate forms, a core differentiation technique for evaluating undefined limits. You will learn to identify all indeterminate forms, convert them to applicable structures, and apply the rule correctly for AP exam questions.

**Prerequisites:** [Evaluate one-sided and two-sided algebraic limits](https://www.owlsprep.com/study/ap-calculus-bc-introduction-to-limits/); [Compute derivatives of all function types, including transcendental functions](https://www.owlsprep.com/study/ap-calculus-bc-basic-derivative-rules/); [Apply the chain rule for differentiation](https://www.owlsprep.com/study/ap-calculus-bc-chain-rule/)

## Learning objectives

- State L'Hopital's rule and its applicability conditions
- Identify all indeterminate forms tested on the AP exam
- Convert non-0/0, non-∞/∞ indeterminate forms to applicable forms
- Apply L'Hopital's rule to correctly evaluate all types of indeterminate limits

## Core L'Hopital's Rule for 0/0 and ∞/∞ Forms

L'Hopital's rule is a differentiation-based technique to evaluate limits of indeterminate forms, where direct substitution gives an expression with no defined limiting value. It applies directly only to the indeterminate forms $\frac{0}{0}$ and $\frac{\infty}{\infty}$; all other indeterminate forms must be rewritten to fit one of these two structures before the rule can be used.

**L'Hopital's 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 $\lim_{x \to a} g(x) = \pm\infty$, $f$ and $g$ are differentiable near $a$ (except possibly at $a$), and $g'(x) \neq 0$ near $a$ (except possibly at $a$), then $\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$. This holds for one-sided limits and limits at infinity.

*Notation:* $\lim_{x \to a} \frac{f(x)}{g(x)}$

*Example:* Applies to $\lim_{x \to 0} \frac{e^x - 1}{x}$ but not to $\lim_{x \to 0} \frac{x}{x+1}$

The rule can be applied repeatedly: if after one application you still get an indeterminate $\frac{0}{0}$ or $\frac{\infty}{\infty}$, you can apply it again as long as all conditions hold.

**Worked example:** Evaluate $\lim_{x \to 0} \frac{e^{2x} - 1 - 2x}{x^2}$

1. Check indeterminacy by direct substitution: numerator $= e^0 - 1 - 2(0) = 0$, denominator $= 0^2 = 0$. This is a valid $\frac{0}{0}$ indeterminate form, so we can apply L'Hopital's rule.
2. Differentiate numerator and denominator separately:
3. $$f'(x) = 2e^{2x} - 2, \quad g'(x) = 2x$$
4. The new limit is $\lim_{x \to 0} \frac{2e^{2x} - 2}{2x} = \frac{0}{0}$, which is still indeterminate, so apply L'Hopital's rule again.
5. Differentiate a second time:
6. $$f''(x) = 4e^{2x}, \quad g''(x) = 2$$
7. Evaluate the new limit:
8. $$\lim_{x \to 0} \frac{4e^{2x}}{2} = \frac{4(1)}{2} = 2$$
9. 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.

## Indeterminate Products: $0 \cdot \infty$

An indeterminate product occurs when one term approaches 0 and the other approaches $\pm\infty$, written $0 \cdot \infty$. The product's limit is indeterminate because 0 pulls the product toward 0 while $\infty$ pulls it toward infinity, so the result can be any finite value, 0, or infinity. To apply L'Hopital's rule, rewrite the product as a fraction by moving one term to the denominator, resulting in either $\frac{0}{0}$ or $\frac{\infty}{\infty}$.

Both conversions are mathematically valid, but one is almost always simpler to differentiate, so choose the conversion that minimizes extra work like the 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 rule. (Moving $\ln(4x)$ to the denominator would result in a more complex differentiation, so this conversion is preferred.)
3. Differentiate numerator and denominator separately:
4. $$\frac{d}{dx}[\ln(4x)] = \frac{1}{x}, \quad \frac{d}{dx}\left[\frac{1}{x}\right] = -\frac{1}{x^2}$$
5. Simplify the ratio of derivatives:
6. $$\frac{1/x}{-1/x^2} = -x$$
7. 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.

## Indeterminate Differences: $\infty - \infty$

Indeterminate differences occur when we have the difference of two terms, both approaching $+\infty$ or both approaching $-\infty$, written $\infty - \infty$. This is indeterminate because the two infinite terms compete, and the result can be 0, any finite number, or $\pm\infty$. To solve this type of limit, 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. The resulting fraction will almost always be a $\frac{0}{0}$ or $\frac{\infty}{\infty}$ indeterminate form suitable for L'Hopital's rule.

**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 terms approach $\pm\infty$, so this is an indeterminate difference $\infty - \infty$.
2. Combine the fractions over a common denominator to get a single fraction:
3. $$\lim_{x \to 1} \frac{\ln x - (x-1)}{(x-1)\ln x}$$
4. Check indeterminacy: numerator at $x=1$ is $\ln 1 - 0 = 0$, denominator is $(0)(\ln 1) = 0$, so we have $\frac{0}{0}$, valid for L'Hopital's rule.
5. Differentiate numerator and denominator:
6. $$\text{Numerator derivative: } \frac{1}{x} - 1 = \frac{1-x}{x}, \quad \text{Denominator derivative: } \ln x + 1 - \frac{1}{x}$$
7. Substitute $x=1$: $\frac{0}{0 + 1 - 1} = \frac{0}{0}$, which is still indeterminate, so apply L'Hopital's rule again.
8. Differentiate a second time:
9. $$\text{Numerator derivative: } -\frac{1}{x^2}, \quad \text{Denominator derivative: } \frac{1}{x} + \frac{1}{x^2}$$
10. Evaluate the limit:
11. $$\lim_{x \to 1} \frac{-1/x^2}{1/x + 1/x^2} = \frac{-1}{1 + 1} = -\frac{1}{2}$$
12. 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.

## 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 the three indeterminate combinations. To solve these, use the natural logarithm to convert the exponential into a product, which you can then convert to $\frac{0}{0}$ or $\frac{\infty}{\infty}$ to apply L'Hopital's rule. The standard 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 take the natural log of both sides:
3. $$\ln L = \lim_{x \to 0^+} x \ln(3x)$$
4. This is an indeterminate product $0 \cdot (-\infty)$. Rewrite as a fraction:
5. $$\lim_{x \to 0^+} \frac{\ln(3x)}{1/x}, \text{ which is } -\infty/\infty$$
6. Differentiate numerator and denominator, then simplify:
7. $$\frac{1/x}{-1/x^2} = -x$$
8. Evaluate the limit: $\ln L = \lim_{x \to 0^+} (-x) = 0$.
9. 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.

## AP-Style Concept Check

**Check your understanding**

Test your understanding of L'Hopital's rule with this AP-style multiple choice question:

1. What is the value of $\lim_{x \to \infty} \frac{5x^3 + 2x - 1}{3x^3 - 4x^2}$?

   - \(0\)
   - \(\frac{5}{3}\)
   - \(\infty\)
   - \(\frac{3}{5}\)

   *Answer:* \(\frac{5}{3}\)

   *Why:* This is an indeterminate $\infty/\infty$ form. Applying L'Hopital's rule three times (or comparing leading terms of the polynomials) gives the result $\frac{5}{3}$.

## Common pitfalls

- **Wrong:** 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 it fails: Students get in the habit of using L'Hopital's for every limit and forget to check the indeterminacy condition first.
  - Correct: Always plug in the limit value first to confirm you have an indeterminate form before applying L'Hopital's.
- **Wrong:** 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 it fails: Confusion between L'Hopital's rule and the derivative quotient rule, since we work with a ratio of functions.
  - Correct: Explicitly label $f(x)$ (numerator) and $g(x)$ (denominator) on scratch paper before differentiating to avoid mixing up rules.
- **Wrong:** Stopping after one application of L'Hopital's when the result is still indeterminate, leaving the answer as $0/0$.
  - Why it fails: Students assume one differentiation is enough and don't check the new limit for indeterminacy.
  - Correct: After each differentiation step, substitute the limit value to check for indeterminacy; apply L'Hopital's again if the result is still indeterminate.
- **Wrong:** Forgetting to exponentiate after using the logarithm for an indeterminate power, leaving the answer as $\ln L = 2$ instead of $L = e^2$.
  - Why it fails: 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: Write "Original limit $L = e^{\ln L}$" explicitly before solving for $\ln L$ to remind yourself of the final step.
- **Wrong:** 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 it fails: L'Hopital's rule only applies to differentiable functions, which sequences are not.
  - Correct: 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.

## Cheatsheet

| Category | Rule/Conversion | Key Notes |
| --- | --- | --- |
| Core L'Hopital's Rule | If $\lim \frac{f}{g} = \frac{0}{0}$ or $\frac{\pm\infty}{\pm\infty}$, then $\lim \frac{f(x)}{g(x)} = \lim \frac{f'(x)}{g'(x)}$ | Holds for $x \to a, x \to a^\pm, x \to \pm\infty$; requires $g'(x) \neq 0$ near $a$ |
| Indeterminate Product $0 \cdot \infty$ | $\lim f g = \lim \frac{f}{1/g}$ or $\lim \frac{g}{1/f}$ | Convert to 0/0 or ∞/∞; choose conversion to simplify differentiation |
| Indeterminate Difference $\infty - \infty$ | Combine to single fraction over common denominator | Use conjugate for radical differences; confirm indeterminate form before applying rule |
| Indeterminate Power $0^0, 1^\infty, \infty^0$ | Let $L = \lim f^g$, $\ln L = \lim g \ln f$, $L = e^{\ln L}$ | Always undo the natural logarithm for the final answer |
| Repeated L'Hopital's | Apply rule multiple times if new limit is still indeterminate | Stop only when you get a determinate finite or infinite limit |
| Determinate Non-Indeterminate Forms | $\frac{0}{\infty} = 0$, $\frac{\infty}{0} = \pm\infty$, $\infty + \infty = \infty$, $0^\infty = 0$ | Never apply L'Hopital's to these; they resolve directly |
| Discrete Sequence Limits | $\lim_{n \to \infty} a_n = \lim_{x \to \infty} f(x)$ where $f(n) = a_n$ | Extend discrete sequences to continuous functions before applying the rule |

## What's next

L'Hopital's rule is a foundational tool for evaluating indeterminate limits that you will use across multiple remaining topics in AP Calculus BC. The most immediate application is evaluating limits for improper integrals, where you will regularly need to resolve indeterminate forms to determine if an integral converges or diverges. You will also use it to compare growth rates of transcendental functions and evaluate limits of infinite sequences for series topics. Mastering the conversion steps for all indeterminate forms now will save you time and avoid common errors on later topics.

- [Analytical Applications of Differentiation Overview](https://www.owlsprep.com/study/ap-calculus-bc-u5-overview/)
- [Mean Value Theorem (MVT)](https://www.owlsprep.com/study/ap-calculus-bc-u5-mean-value-theorem/)
- [Extreme Value Theorem, Global vs Local Extrema, Critical Points](https://www.owlsprep.com/study/ap-calculus-bc-u5-extreme-value-theorem-global-vs/)

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