University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Section 2.2 - Limit of a Function and Limit Laws - Exercises - Page 67: 32


$\lim\limits_{x \to 0}\frac{\frac{1}{x-1} +\frac{1}{x+1}}{x} = -2$

Work Step by Step

Simplify: $\frac{\frac{1}{x-1} +\frac{1}{x+1}}{x} = \frac{(x+1+x-1)}{x(x^{2}-1)} = \frac{2}{(x^{2}-1)}$ Now, $\lim\limits_{x \to 0}\frac{\frac{1}{x-1} +\frac{1}{x+1}}{x} = \lim\limits_{x \to 0} \frac{2}{(x^{2}-1)} = \frac{2}{-1} = -2$
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.