Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.3 - Calculating Limits Using the Limit Laws - 2.3 Exercises - Page 103: 19



Work Step by Step

$\lim\limits_{x\to-2}\frac{x+2}{x^3+8}$ $=\lim\limits_{x\to-2}\frac{x+2}{(x+2)(x^2-2x+4)}$ (remember $(a^3+b^3)=(a+b)(a^2-ab+b^2)$ $=\lim\limits_{x\to-2}\frac{1}{x^2-2x+4}$ (divide both numerator and denominator by $x+2$) $=\frac{1}{(-2)^2-2\times(-2)+4}$ $=\frac{1}{12}$
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