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 - 1.3 Exercises: 52

Answer

$\lim\limits_{x\to 2}\dfrac{x^2+2x-8}{x^2-x-2}=2.$

Work Step by Step

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