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: 51

Answer

$\lim\limits_{x\to -3}\dfrac{x^2+x-6}{x^2-9}=\dfrac{5}{6}.$

Work Step by Step

$\lim\limits_{x\to -3}\dfrac{x^2+x-6}{x^2-9}=\lim\limits_{x\to -3}\dfrac{(x-2)(x+3)}{(x-3)(x+3)}=\lim\limits_{x\to -3}\dfrac{x-2}{x-3}$ $=\dfrac{-3-2}{-3-3}=\dfrac{5}{6}.$
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