University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Section 2.2 - Limit of a Function and Limit Laws - Exercises: 23

Answer

$\lim\limits_{x \to 5}\frac{(x-5)}{x^{2} - 25} = \frac{1}{10} = 0.1$

Work Step by Step

Simplify: $\frac{(x-5)}{x^{2} - 25} = \frac{(x-5)}{(x-5)(x+5)} = \frac{1}{(x+5)}$ Now, $\lim\limits_{x \to 5} \frac{(x-5)}{x^{2} - 25}= \lim\limits_{x \to 5}\frac{1}{(x+5)}$ = $\frac{1}{10}= 0.1$
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