Calculus (3rd Edition)

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

Chapter 2 - Limits - 2.5 Evaluating Limits Algebraically - Exercises - Page 72: 5



Work Step by Step

Numerator at $x=7:\,\,\,x-7=7-7=0$ Denominator at $x=7:\,\,\,x^{2}-49=7^{2}-49=0$ The function has the indeterminate form $\frac{0}{0}$ at $x=7$. Transforming algebraically and canceling the common factor, we have $\frac{x-7}{x^{2}-49}=\frac{x-7}{(x-7)(x+7)}=\frac{1}{x+7}$ Therefore, $\lim\limits_{x \to 7}\frac{x-7}{x^{2}-49}=\lim\limits_{x \to 7}\frac{1}{x+7}=\frac{1}{7+7}=\frac{1}{14}$
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