Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.6 - Limits at Infinity; Horizontal Asymptotes - 2.6 Exercises - Page 139: 51


There are no horizontal asymptotes. $x=5~$ is a vertical asymptote.

Work Step by Step

Horizontal asymptotes: $\lim\limits_{x \to -\infty}\frac{x^3-x}{x^2-6x+5} = -\infty$ $\lim\limits_{x \to \infty}\frac{x^3-x}{x^2-6x+5} = \infty$ There are no horizontal asymptotes. Vertical asymptotes: $\frac{x^3-x}{x^2-6x+5} = \frac{(x)(x-1)(x+1)}{(x-5)(x-1)}$ $x=5~~$ is a vertical asymptote. Although $(x-1)$ is in the denominator, we can see that $(x-1)$ is also in the numerator. Therefore, $~~x=1~~$ is not a vertical asymptote, even though the graph is undefined at this point.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.