# Limits at Infinity Practice Problems with Answers and Solutions

> 12 limits at infinity practice problems with answers and full solutions: rational functions, square roots, exponentials and horizontal asymptotes.

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

$$ \lim_{x\to\infty}\frac{a x^n + \cdots}{b x^n + \cdots} = \frac{a}{b} $$

## At a glance

- **Problems:** 12, with answers and full solutions
- **Level:** Calculus 1 and AP Calculus AB, Unit 1
- **You need:** The fact that $\frac{1}{x^p}\to 0$ as $x\to\infty$ for $p > 0$
- **Main idea:** Divide by the highest power of $x$ in the denominator, then let the small terms go to 0
- **Practice with new numbers:** [Limits at infinity lesson](/learn/calculus-1/limits/limits-at-infinity)

## Key takeaways

- For a rational function, compare degrees: bottom bigger gives $0$, equal degrees give the ratio of leading coefficients, top bigger gives $\pm\infty$.
- With square roots, remember $\sqrt{x^2} = |x|$, which is $-x$ when $x\to -\infty$. That sign flip changes the answer.
- An $\infty - \infty$ form with a square root usually needs the conjugate before you can see the limit.

These **limits at infinity practice problems** start with rational functions and move on to square roots, the $\infty - \infty$ form, exponentials and horizontal asymptotes. Work each one before you check. The answers come right after the list, and the full solutions follow.

The shortcut for rational functions, where $n$ is the degree of the top and $m$ the degree of the bottom:

$$\lim_{x\to\pm\infty}\frac{\text{top}}{\text{bottom}} = \begin{cases} 0 & n < m \\ \dfrac{\text{leading coefficient of top}}{\text{leading coefficient of bottom}} & n = m \\ \pm\infty & n > m \end{cases}$$

For the reasoning behind it, read the [limits at infinity guide](/blog/limits-at-infinity).

## Problems

**Rational functions**

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

**Square roots**

5. $\displaystyle\lim_{x\to\infty}\frac{\sqrt{9x^2 + 1}}{2x + 5}$
6. $\displaystyle\lim_{x\to -\infty}\frac{\sqrt{9x^2 + 1}}{2x + 5}$
7. $\displaystyle\lim_{x\to\infty}\left(\sqrt{x^2 + 6x} - x\right)$

**Trig and exponentials**

8. $\displaystyle\lim_{x\to\infty}\frac{\sin x}{x}$
9. $\displaystyle\lim_{x\to\infty}\frac{2e^x + 3}{e^x - 1}$
10. $\displaystyle\lim_{x\to -\infty}\frac{2e^x + 3}{e^x - 1}$
11. $\displaystyle\lim_{x\to\infty}\arctan(x^2)$

**Horizontal asymptotes**

12. Find all horizontal asymptotes of $f(x) = \dfrac{5x - 2}{\sqrt{x^2 + 4}}$.

## Answers

| Problem | Answer |
| --- | --- |
| 1 | $\dfrac{3}{5}$ |
| 2 | $0$ |
| 3 | $\infty$ |
| 4 | $-\dfrac{1}{3}$ |
| 5 | $\dfrac{3}{2}$ |
| 6 | $-\dfrac{3}{2}$ |
| 7 | $3$ |
| 8 | $0$ |
| 9 | $2$ |
| 10 | $-3$ |
| 11 | $\dfrac{\pi}{2}$ |
| 12 | $y = 5$ (as $x\to\infty$) and $y = -5$ (as $x\to -\infty$) |

## Step-by-step solutions

**1.** The degrees match, so divide top and bottom by $x^2$:

$$\frac{3 + \frac{2}{x}}{5 - \frac{1}{x^2}} \;\to\; \frac{3 + 0}{5 - 0} = \frac{3}{5}$$

**2.** The bottom has the higher degree. Divide by $x^2$:

$$\frac{\frac{4}{x} + \frac{1}{x^2}}{1 + \frac{3}{x^2}} \;\to\; \frac{0}{1} = 0$$

**3.** The top has the higher degree. Divide by $x^2$:

$$\frac{x - \frac{2}{x^2}}{4 + \frac{1}{x}}$$

The top grows without bound while the bottom approaches $4$, so the limit is $\infty$.

**4.** Equal degrees again. Divide by $x^3$:

$$\frac{2 - \frac{1}{x^2}}{\frac{1}{x^3} - 6} \;\to\; \frac{2}{-6} = -\frac{1}{3}$$

The direction $-\infty$ does not change the answer here, because every leftover term still goes to $0$.

**5.** Factor $x^2$ out of the root. For $x > 0$, $\sqrt{x^2} = x$:

$$\frac{x\sqrt{9 + \frac{1}{x^2}}}{x\left(2 + \frac{5}{x}\right)} = \frac{\sqrt{9 + \frac{1}{x^2}}}{2 + \frac{5}{x}} \;\to\; \frac{3}{2}$$

**6.** Same function, but now $x < 0$, so $\sqrt{x^2} = |x| = -x$:

$$\frac{-x\sqrt{9 + \frac{1}{x^2}}}{x\left(2 + \frac{5}{x}\right)} = -\frac{\sqrt{9 + \frac{1}{x^2}}}{2 + \frac{5}{x}} \;\to\; -\frac{3}{2}$$

**7.** This is an $\infty - \infty$ form. Multiply by the conjugate over itself:

$$\frac{(x^2 + 6x) - x^2}{\sqrt{x^2 + 6x} + x} = \frac{6x}{\sqrt{x^2 + 6x} + x}$$

Divide top and bottom by $x$:

$$\frac{6}{\sqrt{1 + \frac{6}{x}} + 1} \;\to\; \frac{6}{1 + 1} = 3$$

**8.** Use the [squeeze theorem](/theorems/squeeze-theorem). Since $-1 \le \sin x \le 1$, for $x > 0$:

$$-\frac{1}{x} \le \frac{\sin x}{x} \le \frac{1}{x}$$

Both outer functions go to $0$, so the limit is $0$.

**9.** As $x\to\infty$, $e^x$ is the dominant term. Divide by $e^x$:

$$\frac{2 + 3e^{-x}}{1 - e^{-x}} \;\to\; \frac{2 + 0}{1 - 0} = 2$$

**10.** As $x\to -\infty$, $e^x\to 0$, so just substitute that value:

$$\frac{2(0) + 3}{0 - 1} = -3$$

**11.** As $x\to\infty$, $x^2\to\infty$, and $\arctan u\to\frac{\pi}{2}$ as $u\to\infty$. So the limit is $\frac{\pi}{2}$.

**12.** Factor $\sqrt{x^2} = |x|$ out of the root: $\sqrt{x^2 + 4} = |x|\sqrt{1 + \frac{4}{x^2}}$.

As $x\to\infty$, $|x| = x$:

$$\frac{5x - 2}{x\sqrt{1 + \frac{4}{x^2}}} = \frac{5 - \frac{2}{x}}{\sqrt{1 + \frac{4}{x^2}}} \;\to\; 5$$

As $x\to -\infty$, $|x| = -x$, which flips the sign and gives $-5$. The graph has two horizontal asymptotes: $y = 5$ on the right and $y = -5$ on the left.

## Common mistakes on these problems

- **Writing $\frac{\infty}{\infty} = 1$.** In problem 1 the top and bottom both grow, but the answer is $\frac{3}{5}$. Divide by the highest power to see what really happens.
- **Forgetting $\sqrt{x^2} = |x|$.** Problems 6 and 12 give the wrong sign if you treat $\sqrt{x^2}$ as $x$ when $x$ is negative.
- **Writing $\infty - \infty = 0$.** Problem 7 looks like it should be $0$, but the conjugate shows the limit is $3$.
- **Assuming both directions match.** Problems 9 and 10 use the same function and get $2$ and $-3$. Exponentials behave very differently on each side.

## Keep practicing

These 12 problems use fixed numbers. For unlimited practice with new numbers every time and instant feedback, try the [limits at infinity lesson](/learn/calculus-1/limits/limits-at-infinity), then the mixed [limits practice](/learn/calculus-1/limits/limits-practice). If a limit at infinity turns into $\frac{\infty}{\infty}$ with exponentials and logs mixed together, the [L'Hôpital's rule practice problems](/practice-problems/lhopitals-rule) show the faster method.

## Frequently asked questions

### How do you find a limit at infinity?

Divide every term by the highest power of $x$ in the denominator. Terms like $\frac{3}{x}$ and $\frac{1}{x^2}$ go to $0$, and whatever is left is the limit. For rational functions you can skip straight to comparing the degrees of the top and bottom.

### How are limits at infinity related to horizontal asymptotes?

If $\lim_{x\to\infty} f(x) = L$ or $\lim_{x\to -\infty} f(x) = L$, then the line $y = L$ is a horizontal asymptote. A function can have two different horizontal asymptotes, one on each side, as in problem 12.

### Why is the limit different as x goes to negative infinity?

Usually because of a square root or an exponential. $\sqrt{x^2} = |x|$ equals $-x$ for negative $x$, and $e^x\to 0$ as $x\to -\infty$ but grows without bound as $x\to\infty$. Polynomials and rational functions without roots often give the same finite limit on both sides.