Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 2: Limits and Continuity - Section 2.2 - Limit of a Function and Limit Laws - Exercises 2.2 - Page 57: 32

Answer

$-1$

Work Step by Step

We want to find $\lim\limits_{x \to 0}\dfrac{\frac{1}{x-1}+\frac{1}{x+1}}{x}$, but we can't use the quotient rule for limits because the limit of the denominator as $x$ approaches 0 is zero, and if we try to substitute $x=0$ directly, we get zero in both the numerator and the denominator. But, note that $\dfrac{\frac{1}{x-1}+\frac{1}{x+1}}{x}=\left(\dfrac{\frac{1}{x-1}+\frac{1}{x+1}}{x}\right)\left(\dfrac{(x-1)(x+1)}{(x-1)(x+1)}\right)=\dfrac{x+1+x-1}{x(x-1)(x+1)}=\dfrac{2x}{x(x-1)(x+1)}=\dfrac{2}{(x-1)(x+1)}$ for all $x \neq 0.$ Thus $\lim\limits_{x \to 0}\dfrac{\frac{1}{x-1}+\frac{1}{x+1}}{x}=\lim\limits_{x \to 0}\dfrac{2}{(x-1)(x+1)}=\dfrac{2}{(0-1)(0+1)}=\dfrac{2}{(-1)(1)}=-1.$
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