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: 60



Work Step by Step

Here, we have $f(x)=\frac{1}{x};x=-2$ and $f(x+h)=\dfrac{1}{x+h}$ Now, $\lim\limits_{h\to0}\dfrac{f(x+h)-f(x)}{h}=\lim\limits_{h\to0}\dfrac{(\dfrac{1}{x+h})-(\dfrac{1}{x})}{h}=\lim\limits_{h\to0}\dfrac{-1}{x(x+h)}$ Plug $x=-2$, we get $\lim\limits_{h\to0}\dfrac{-1}{x(x+h)}=\dfrac{-1}{(-2)(-2+0)}$ or, $\dfrac{-1}{(-2)(-2)}=-\dfrac{1}{4}$
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