# L'Hôpital's Rule Practice Problems with Answers and Solutions

> 12 L'Hôpital's rule practice problems with answers and full solutions, covering 0/0, ∞/∞, 0·∞, ∞−∞ and exponent forms like 1^∞ and 0^0.

Source: https://practicecalculus.com/practice-problems/lhopitals-rule
Updated: 2026-11-06

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

## At a glance

- **Problems:** 12, with answers and full solutions
- **Level:** Calculus 1 and AP Calculus AB, Unit 4
- **You need:** The basic derivative rules and [limits at infinity](/practice-problems/limits-at-infinity)
- **Main idea:** Check for 0/0 or ∞/∞, then differentiate the top and the bottom separately
- **Practice with new numbers:** [L'Hôpital's rule lesson](/learn/calculus-1/applications-of-derivatives/lhopitals-rule)

## Key takeaways

- Only use the rule on $\frac{0}{0}$ or $\frac{\infty}{\infty}$. Check the form every time, including before a second application.
- Differentiate the top and the bottom separately. Do not use the quotient rule.
- Rewrite $0\cdot\infty$ and $\infty-\infty$ as a single fraction first, and handle $1^\infty$, $0^0$ and $\infty^0$ by taking the natural log.

These **L'Hôpital's rule practice problems** cover all seven indeterminate forms: $\frac{0}{0}$, $\frac{\infty}{\infty}$, $0\cdot\infty$, $\infty - \infty$, $0^0$, $1^\infty$ and $\infty^0$. The last problem is a trap where the rule does not apply. The answers come right after the list, then the full solutions.

The rule:

$$\lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{x\to a}\frac{f'(x)}{g'(x)} \quad\text{if the first limit is } \tfrac{0}{0} \text{ or } \tfrac{\infty}{\infty}$$

For the full statement and why it works, see the [L'Hôpital's rule page](/rules/lhopitals-rule).

## Problems

**0/0 forms**

1. $\displaystyle\lim_{x\to 0}\frac{e^{2x} - 1}{\sin x}$
2. $\displaystyle\lim_{x\to 1}\frac{\ln x}{x - 1}$
3. $\displaystyle\lim_{x\to 0}\frac{x - \sin x}{x^3}$

**∞/∞ forms**

4. $\displaystyle\lim_{x\to\infty}\frac{x^2}{e^x}$
5. $\displaystyle\lim_{x\to\infty}\frac{\ln x}{\sqrt{x}}$

**Products and differences**

6. $\displaystyle\lim_{x\to 0^+} x\ln x$
7. $\displaystyle\lim_{x\to\infty} x\sin\!\left(\frac{1}{x}\right)$
8. $\displaystyle\lim_{x\to 0}\left(\frac{1}{x} - \frac{1}{e^x - 1}\right)$

**Exponent forms**

9. $\displaystyle\lim_{x\to 0^+} x^x$
10. $\displaystyle\lim_{x\to\infty} x^{1/x}$
11. $\displaystyle\lim_{x\to\infty}\left(1 + \frac{3}{x}\right)^x$

**Check before you start**

12. $\displaystyle\lim_{x\to 0}\frac{x^2 + 3}{x + 1}$

## Answers

| Problem | Answer |
| --- | --- |
| 1 | $2$ |
| 2 | $1$ |
| 3 | $\dfrac{1}{6}$ |
| 4 | $0$ |
| 5 | $0$ |
| 6 | $0$ |
| 7 | $1$ |
| 8 | $\dfrac{1}{2}$ |
| 9 | $1$ |
| 10 | $1$ |
| 11 | $e^3$ |
| 12 | $3$ (L'Hôpital does not apply) |

## Step-by-step solutions

**1.** Substitution gives $\frac{0}{0}$. Differentiate top and bottom:

$$\lim_{x\to 0}\frac{2e^{2x}}{\cos x} = \frac{2}{1} = 2$$

**2.** Substitution gives $\frac{\ln 1}{0} = \frac{0}{0}$:

$$\lim_{x\to 1}\frac{1/x}{1} = 1$$

**3.** Each round still gives $\frac{0}{0}$, so apply the rule three times:

$$\lim_{x\to 0}\frac{x - \sin x}{x^3} = \lim_{x\to 0}\frac{1 - \cos x}{3x^2} = \lim_{x\to 0}\frac{\sin x}{6x} = \lim_{x\to 0}\frac{\cos x}{6} = \frac{1}{6}$$

**4.** This is $\frac{\infty}{\infty}$. Two rounds:

$$\lim_{x\to\infty}\frac{2x}{e^x} = \lim_{x\to\infty}\frac{2}{e^x} = 0$$

Exponentials beat every power of $x$.

**5.** Again $\frac{\infty}{\infty}$:

$$\lim_{x\to\infty}\frac{1/x}{\frac{1}{2\sqrt{x}}} = \lim_{x\to\infty}\frac{2\sqrt{x}}{x} = \lim_{x\to\infty}\frac{2}{\sqrt{x}} = 0$$

Logs lose to every positive power of $x$.

**6.** This is $0\cdot(-\infty)$. Move $x$ to the bottom as $\frac{1}{x}$ to get $\frac{-\infty}{\infty}$:

$$\lim_{x\to 0^+}\frac{\ln x}{1/x} = \lim_{x\to 0^+}\frac{1/x}{-1/x^2} = \lim_{x\to 0^+}(-x) = 0$$

**7.** This is $\infty\cdot 0$. Rewrite as $\frac{\sin(1/x)}{1/x}$, which is $\frac{0}{0}$. Let $t = \frac{1}{x}$, so $t\to 0^+$:

$$\lim_{t\to 0^+}\frac{\sin t}{t} = \lim_{t\to 0^+}\frac{\cos t}{1} = 1$$

**8.** This is $\infty - \infty$. Combine into one fraction:

$$\frac{1}{x} - \frac{1}{e^x - 1} = \frac{e^x - 1 - x}{x(e^x - 1)}$$

That is $\frac{0}{0}$. Differentiate, and it is still $\frac{0}{0}$, so differentiate again:

$$\lim_{x\to 0}\frac{e^x - 1}{e^x - 1 + xe^x} = \lim_{x\to 0}\frac{e^x}{2e^x + xe^x} = \frac{1}{2}$$

**9.** This is $0^0$. Let $L$ be the limit and take logs. Problem 6 already found this limit:

$$\ln L = \lim_{x\to 0^+} x\ln x = 0 \quad\Rightarrow\quad L = e^0 = 1$$

**10.** This is $\infty^0$. Take logs:

$$\ln L = \lim_{x\to\infty}\frac{\ln x}{x} = \lim_{x\to\infty}\frac{1/x}{1} = 0 \quad\Rightarrow\quad L = 1$$

**11.** This is $1^\infty$. Take logs, then write the product as a $\frac{0}{0}$ fraction:

$$\ln L = \lim_{x\to\infty} x\ln\!\left(1 + \frac{3}{x}\right) = \lim_{x\to\infty}\frac{\ln\left(1 + \frac{3}{x}\right)}{1/x}$$

Differentiate the top, $\frac{1}{1 + 3/x}\cdot\left(-\frac{3}{x^2}\right)$, and the bottom, $-\frac{1}{x^2}$:

$$\ln L = \lim_{x\to\infty}\frac{3}{1 + \frac{3}{x}} = 3 \quad\Rightarrow\quad L = e^3$$

**12.** Substitute first: $\frac{0 + 3}{0 + 1} = 3$. That is a real number, so the limit is $3$ and you are done. Using L'Hôpital here would give $\frac{2x}{1}\to 0$, which is wrong, because the rule only works on indeterminate forms.

## Common mistakes on these problems

- **Using the quotient rule.** L'Hôpital means $\frac{f'}{g'}$, not $\left(\frac{f}{g}\right)'$. Differentiate the top and bottom on their own.
- **Skipping the check.** Problem 12 shows what happens: the rule gives a wrong answer when the form is not indeterminate.
- **Forgetting to undo the log.** In problems 9 to 11 you find $\ln L$ first. The answer to problem 11 is $e^3$, not $3$.
- **Applying the rule to a product directly.** In problem 6 you must rewrite $x\ln x$ as a fraction before differentiating.

## Keep practicing

These 12 problems use fixed numbers. For unlimited practice with new limits every time and instant feedback, try the [L'Hôpital's rule lesson](/learn/calculus-1/applications-of-derivatives/lhopitals-rule). For more worked examples and the reasoning behind each indeterminate form, read the [L'Hôpital's rule guide](/blog/lhopitals-rule). To review the algebra methods first, go back to the [evaluating limits practice problems](/practice-problems/evaluating-limits).

## Frequently asked questions

### When can you use L'Hôpital's rule?

Only when direct substitution gives $\frac{0}{0}$ or $\frac{\infty}{\infty}$, and the top and bottom are differentiable near the point. Other indeterminate forms must be rewritten into one of these two forms first.

### Can you use L'Hôpital's rule more than once?

Yes, as long as the new limit is still $\frac{0}{0}$ or $\frac{\infty}{\infty}$. Problem 3 below needs three rounds. Check the form before each round, because once it stops being indeterminate you must substitute.

### How do you use L'Hôpital's rule on 1 to the infinity?

Call the limit $L$ and take the natural log: $\ln L = \lim g(x)\ln f(x)$. That turns the power into a product, which you rewrite as a fraction and solve with L'Hôpital. Then $L = e^{\ln L}$. Problem 11 shows this and gives $e^3$.