Answer
$\lim\limits_{x\to 2}\dfrac{x^2+2x-8}{x^2-x-2}=2.$
Work Step by Step
$\lim\limits_{x\to 2}\dfrac{x^2+2x-8}{x^2-x-2}=\lim\limits_{x\to 2}\dfrac{(x+4)(x-2)}{(x+1)(x-2)}=\lim\limits_{x\to 2}\dfrac{x+4}{x+1}$
$=\dfrac{2+4}{2+1}=2.$
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 - 1.3 Exercises - Page 67: 52
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