Answer
$\lim\limits_{x\to 4}\dfrac{x-4}{x^2-16}=\dfrac{1}{8}.$
Work Step by Step
$\lim\limits_{x\to 4}\dfrac{x-4}{x^2-16}=\lim\limits_{x\to 4}\dfrac{(x-4)}{(x+4)(x-4)}=\lim\limits_{x\to 4}\dfrac{1}{x+4}=\dfrac{1}{4+4}=\dfrac{1}{8}.$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
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 - Page 67: 49
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