Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 2 - Limits - 2.2 Limits: A Numerical and Graphical Approach - Exercises - Page 54: 27



Work Step by Step

Evaluating the function at -3, we get it of the form $\frac{0}{0}$. We try to rewrite the function cancelling the factors which are causing the limit to be of the form 0/0. Hence $\lim\limits_{x \to -3}\frac{x+3}{x^{2}+x-6}=\lim\limits_{x \to -3}\frac{x+3}{(x+3)(x-2)}=\lim\limits_{x \to -3}\frac{1}{x-2}= \frac{1}{-3-2}= -\frac{1}{5}$
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