Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 1 - Functions and Limits - 1.6 Calculating Limits Using the Limit Laws - 1.6 Exercises - Page 70: 32

Answer

$-\frac{2}{x^3}$

Work Step by Step

$\lim\limits_{h \to 0}\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}=\lim\limits_{h \to 0}\frac{\frac{x^2}{x^2(x+h)^2}-\frac{(x+h)^2}{x^2(x+h)^2}}{h}=\lim\limits_{h \to 0}\frac{x^2-(x+h)^2}{x^2(x+h)^2h}=\lim\limits_{h \to 0}\frac{x^2-(x^2+2xh+h^2)}{x^2(x+h^2)h}=\lim\limits_{h \to 0}\frac{-2xh-h^2}{x^2(x+h)^2h}=\lim\limits_{h \to 0}\frac{-2x-h}{x^2(x+h)^2}=\frac{-2x-(0)}{x^2(x+(0))^2}=\frac{-2x}{x^4}=-\frac{2}{x^3}$
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