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: 49

Answer

$\lim\limits_{x\to 4}\dfrac{x-4}{x^2-16}=\dfrac{1}{8}.$

Work Step by Step

$\lim\limits_{x\to 4}\dfrac{x-4}{x^2-16}=\lim\limits_{x\to 4}\dfrac{(x-4)}{(x+4)(x-4)}=\lim\limits_{x\to 4}\dfrac{1}{x+4}=\dfrac{1}{4+4}=\dfrac{1}{8}.$
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