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 - 1.4 Exercises: 10

Answer

$\lim\limits_{x\to4^+}\dfrac{4-x}{x^2-16}=-\dfrac{1}{8}.$

Work Step by Step

$\lim\limits_{x\to4^+}\dfrac{4-x}{x^2-16}=\lim\limits_{x\to4^+}\dfrac{-(x-4)}{(x-4)(x+4)}$ $=\lim\limits_{x\to4^+}\dfrac{-1}{x+4}=-\dfrac{1}{4+4}=-\dfrac{1}{8}.$
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