Answer
$\lim\limits_{x\to4^+}\dfrac{4-x}{x^2-16}=-\dfrac{1}{8}.$
Work Step by Step
$\lim\limits_{x\to4^+}\dfrac{4-x}{x^2-16}=\lim\limits_{x\to4^+}\dfrac{-(x-4)}{(x-4)(x+4)}$
$=\lim\limits_{x\to4^+}\dfrac{-1}{x+4}=-\dfrac{1}{4+4}=-\dfrac{1}{8}.$
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 - 1.4 Exercises - Page 79: 10
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