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

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:

For the full statement and why it works, see the L'Hôpital's rule page.

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

Step-by-step solutions

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

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

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

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

Exponentials beat every power of x.

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

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}:

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^+:

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

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

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

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

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

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

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. For more worked examples and the reasoning behind each indeterminate form, read the L'Hôpital's rule guide. To review the algebra methods first, go back to the evaluating limits practice problems.