Answer
$\lim\limits_{x\to8^+}\dfrac{1}{x+8}=\dfrac{1}{16}.$
Work Step by Step
Using direct substitution:
$\lim\limits_{x\to8^+}\dfrac{1}{x+8}=\dfrac{1}{8+8}=\dfrac{1}{16}.$
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: 7
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