Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 1 - Functions and Limits - 1.6 Calculating Limits Using the Limit Laws - 1.6 Exercises: 19

Answer

$\frac{1}{12}$

Work Step by Step

$\lim\limits_{x \to -2}\frac{x+2}{x^3+8}=\frac{(-2)+2}{(-2)^3+8}=\frac{0}{0}=$indet so we have to do algebra $\lim\limits_{x \to -2}\frac{x+2}{x^3+8}=\lim\limits_{x \to -2}\frac{x+2}{(x+2)(x^2-2x+4)}=\lim\limits_{x \to -2}\frac{1}{x^2-2x+4}=\frac{1}{(-2)^2-2(-2)+4}=\frac{1}{12}$
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