# Evaluating Limits Practice Problems with Answers and Solutions

> 12 limit practice problems with answers and step-by-step solutions: direct substitution, factoring 0/0, conjugates, trig limits and one-sided limits.

Source: https://practicecalculus.com/practice-problems/evaluating-limits
Updated: 2026-11-06

$$ \lim_{x\to a} f(x) = f(a) \text{ when } f \text{ is continuous at } a $$

## At a glance

- **Problems:** 12, with answers and full solutions
- **Level:** Calculus 1 and AP Calculus AB, Unit 1
- **You need:** Factoring, fractions and the [limit laws](/rules/limit-laws)
- **Main idea:** Plug in first. If you get 0/0, simplify and plug in again
- **Practice with new numbers:** [Limits lessons](/learn/calculus-1/limits/direct-substitution)

## Key takeaways

- Always try direct substitution first. A real number means you are done.
- A result of $\frac{0}{0}$ means "simplify first": factor, cancel, use a conjugate or combine fractions, then substitute again.
- A nonzero number over $0$ means the limit is infinite or does not exist, so check both one-sided limits.

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:

$$\text{substitute} \;\to\; \text{if } \tfrac{0}{0},\ \text{simplify} \;\to\; \text{substitute again}$$

For a refresher on why this works, read [how to evaluate limits](/blog/how-to-evaluate-limits) and the [direct substitution](/rules/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

| Problem | Answer |
| --- | --- |
| 1 | $3$ |
| 2 | $2$ |
| 3 | $\dfrac{3}{2}$ |
| 4 | $6$ |
| 5 | $-\dfrac{1}{4}$ |
| 6 | $3$ |
| 7 | $\dfrac{1}{6}$ |
| 8 | $-\dfrac{1}{16}$ |
| 9 | $5$ |
| 10 | $\dfrac{1}{2}$ |
| 11 | $+\infty$ |
| 12 | Does not exist (left limit $-1$, right limit $1$) |

## Step-by-step solutions

**1.** Polynomials are continuous everywhere, so substitute:

$$3(2)^2 - 5(2) + 1 = 12 - 10 + 1 = 3$$

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

$$\frac{e^0 + \cos 0}{0 + 1} = \frac{1 + 1}{1} = 2$$

**3.** Again nothing breaks at $x = 3$:

$$\frac{\sqrt{3 + 6}}{3 - 1} = \frac{\sqrt{9}}{2} = \frac{3}{2}$$

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

$$\frac{(x-3)(x+3)}{x-3} = x + 3 \;\to\; 3 + 3 = 6$$

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

$$\frac{(x+2)(x+3)}{(x+2)(x-2)} = \frac{x+3}{x-2} \;\to\; \frac{-2+3}{-2-2} = -\frac{1}{4}$$

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

$$\frac{(x-1)(x^2+x+1)}{x-1} = x^2 + x + 1 \;\to\; 1 + 1 + 1 = 3$$

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

$$\frac{(x + 9) - 9}{x\left(\sqrt{x+9} + 3\right)} = \frac{1}{\sqrt{x+9} + 3} \;\to\; \frac{1}{3 + 3} = \frac{1}{6}$$

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

$$\frac{-(x-4)}{4x(x-4)} = -\frac{1}{4x} \;\to\; -\frac{1}{16}$$

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

$$\frac{\sin(5x)}{x} = 5\cdot\frac{\sin(5x)}{5x} \;\to\; 5\cdot 1 = 5$$

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

$$\frac{\sin^2 x}{x^2(1 + \cos x)} = \left(\frac{\sin x}{x}\right)^2\cdot\frac{1}{1 + \cos x} \;\to\; 1\cdot\frac{1}{2} = \frac{1}{2}$$

**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](/learn/calculus-1/limits/direct-substitution), then [factoring 0/0 forms](/learn/calculus-1/limits/factoring-zero-over-zero) and the [squeeze theorem](/learn/calculus-1/limits/squeeze-theorem) lesson. When you are ready for limits as $x\to\infty$, move on to the [limits at infinity practice problems](/practice-problems/limits-at-infinity).

## Frequently asked questions

### What is the first step in evaluating a limit?

Substitute the value into the function. If the result is a real number, that is the limit. Only when you get a form like $\frac{0}{0}$ or a nonzero number over zero do you need more work.

### What does 0/0 mean when evaluating a limit?

It is an indeterminate form. It does not mean the limit is 0 or undefined. It means the expression needs to be simplified, usually by factoring and cancelling a common factor, before you substitute again.

### When does a limit not exist?

A limit does not exist when the left and right limits are different, as with $\frac{|x|}{x}$ at $0$, when the function grows without bound, or when it oscillates forever, as with $\sin\frac{1}{x}$ near $0$.

### When should I use the conjugate to evaluate a limit?

Use it when a square root appears in a sum or difference and substitution gives $\frac{0}{0}$, as in $\frac{\sqrt{x+9}-3}{x}$. Multiplying by the conjugate clears the root from the top so a factor of $x$ can cancel.