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: 32

Answer

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

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