Evaluating Limits Practice Problems with Answers and Solutions

These evaluating limits practice problems cover every method you need for finite limits: direct substitution, factoring, conjugates, combining fractions, the special trig limits and one-sided limits. Try each problem first. The answers come right after the list, then the full solutions.

The strategy for every problem:

For a refresher on why this works, read how to evaluate limits and the direct substitution rule.

Problems

Direct substitution

1. \displaystyle\lim_{x\to 2}\,(3x^2 - 5x + 1)
2. \displaystyle\lim_{x\to 0}\,\frac{e^x + \cos x}{x + 1}
3. \displaystyle\lim_{x\to 3}\,\frac{\sqrt{x + 6}}{x - 1}

Factoring a 0/0 form

4. \displaystyle\lim_{x\to 3}\,\frac{x^2 - 9}{x - 3}
5. \displaystyle\lim_{x\to -2}\,\frac{x^2 + 5x + 6}{x^2 - 4}
6. \displaystyle\lim_{x\to 1}\,\frac{x^3 - 1}{x - 1}

Conjugates and fractions

7. \displaystyle\lim_{x\to 0}\,\frac{\sqrt{x + 9} - 3}{x}
8. \displaystyle\lim_{x\to 4}\,\frac{\frac{1}{x} - \frac{1}{4}}{x - 4}

Trig limits

9. \displaystyle\lim_{x\to 0}\,\frac{\sin(5x)}{x}
10. \displaystyle\lim_{x\to 0}\,\frac{1 - \cos x}{x^2}

One-sided limits

11. \displaystyle\lim_{x\to 2^+}\,\frac{x + 1}{x - 2}
12. \displaystyle\lim_{x\to 0}\,\frac{|x|}{x}

Answers

Step-by-step solutions

1. Polynomials are continuous everywhere, so substitute:

2. The bottom is not zero at x = 0, so substitute:

3. Again nothing breaks at x = 3:

4. Substitution gives \frac{0}{0}. Factor the difference of squares and cancel:

5. Substitution gives \frac{4 - 10 + 6}{4 - 4} = \frac{0}{0}. Factor both parts:

6. Use the difference of cubes, x^3 - 1 = (x-1)(x^2 + x + 1):

7. Substitution gives \frac{0}{0}. Multiply top and bottom by the conjugate \sqrt{x+9} + 3:

8. Combine the fractions on top: \frac{1}{x} - \frac{1}{4} = \frac{4 - x}{4x}. Since 4 - x = -(x - 4):

9. Use \displaystyle\lim_{\theta\to 0}\frac{\sin\theta}{\theta} = 1 with \theta = 5x. Multiply and divide by 5:

10. Multiply by \frac{1 + \cos x}{1 + \cos x} and use 1 - \cos^2 x = \sin^2 x:

11. Substitution gives \frac{3}{0}, so the limit is infinite. For x slightly bigger than 2, the top is near 3 and the bottom is a small positive number, so the fraction is large and positive: the limit is +\infty. There is a vertical asymptote at x = 2.

12. For x > 0, |x| = x, so the fraction is 1. For x < 0, |x| = -x, so the fraction is -1. The one-sided limits are 1 and -1. They do not match, so the two-sided limit does not exist.

Common mistakes on these problems

• Stopping at 0/0. In problem 4, \frac{0}{0} is not an answer. It tells you to factor first.
• Cancelling terms instead of factors. In problem 5 you can cancel (x+2) because it multiplies the whole top and bottom. You can never cancel the x^2 from x^2 + 5x + 6 and x^2 - 4.
• Writing \frac{\sin 5x}{x} = \sin 5. You cannot cancel the x inside a sine. Problem 9 needs the special limit \frac{\sin\theta}{\theta}\to 1.
• Ignoring the side. In problem 11 the side matters. From the left, the same fraction goes to -\infty, so only the right-hand limit is +\infty.

Keep practicing

These 12 problems use fixed numbers. For unlimited practice with fresh numbers and instant feedback on each step, try direct substitution, then factoring 0/0 forms and the squeeze theorem lesson. When you are ready for limits as x\to\infty, move on to the limits at infinity practice problems.