Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 1 - Limits and Their Properties - 1.3 Exercises: 88

Answer

$\lim\limits_{\Delta x\to0}\dfrac{\dfrac{1}{(x+\Delta x)^2}-\dfrac{1}{x^2}}{\Delta x}=-\dfrac{2}{x^3}$

Work Step by Step

$\lim\limits_{\Delta x\to0}\dfrac{\dfrac{1}{(x+\Delta x)^2}-\dfrac{1}{x^2}}{\Delta x}$ $=\lim\limits_{\Delta x\to0}\dfrac{x^2-(x^2+2x(\Delta x)+(\Delta x)^2)}{\Delta x(x+\Delta x)^2(x)^2}$ $=\lim\limits_{\Delta x\to0}\dfrac{\Delta x(-2x-\Delta x)}{\Delta x(x+\Delta x)^2(x)^2}$ $=\lim\limits_{\Delta x\to0}\dfrac{-\Delta x-2x}{(x+\Delta x)^2(x)^2}$ $=\dfrac{0-2x}{(x+0)^2(x)^2}=-\dfrac{2}{x^3}.$
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