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: 87

Answer

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

Work Step by Step

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