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 - 1.5 Exercises: 37

Answer

$\lim\limits_{x\to-3^-}\dfrac{x+3}{x^2+x-6}=-\dfrac{1}{5}.$

Work Step by Step

$\lim\limits_{x\to-3^-}\dfrac{x+3}{x^2+x-6}=\lim\limits_{x\to-3^-}\dfrac{(x+3)}{(x+3)(x-2)}=\lim\limits_{x\to-3^-}\dfrac{1}{x-2}$ $=\dfrac{1}{-3^--2}=-\dfrac{1}{5}.$
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