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Limits and continuity

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.

Limits and continuity Updated 4 min read

At a glance

\lim_{x\to\infty}\frac{a x^n + \cdots}{b x^n + \cdots} = \frac{a}{b}
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
On this page
  1. Problems
  2. Answers
  3. Step-by-step solutions
  4. Common mistakes on these problems
  5. Keep practicing

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.

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

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

Trig and exponentials

  1. \displaystyle\lim_{x\to\infty}\frac{\sin x}{x}
  2. \displaystyle\lim_{x\to\infty}\frac{2e^x + 3}{e^x - 1}
  3. \displaystyle\lim_{x\to -\infty}\frac{2e^x + 3}{e^x - 1}
  4. \displaystyle\lim_{x\to\infty}\arctan(x^2)

Horizontal asymptotes

  1. 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. 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, then the mixed 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 show the faster method.

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Limits at infinity

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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.

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