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: 42

Answer

$\lim\limits_{x \to 4}\frac{4-x}{5 - \sqrt (x^{2}+9)} = \frac{5}{4} = 1.25$

Work Step by Step

$\frac{4-x}{5 - \sqrt (x^{2}+9)} = \frac{(4-x)(5 + \sqrt (x^{2}+9))}{(5 - \sqrt (x^{2}+9))(5 + \sqrt (x^{2}+9))} = \frac{(4-x)(5 + \sqrt (x^{2}+9))}{25 - (x^{2}+9)} = = \frac{(4-x)(5 + \sqrt (x^{2}+9))}{(16-x^{2})} = \frac{(5 + \sqrt (x^{2}+9))}{(4+x)}$ Now, $\lim\limits_{x \to 4}\frac{4-x}{5 - \sqrt (x^{2}+9)} = \lim\limits_{x \to 4}\frac{(5 + \sqrt (x^{2}+9))}{(4+x)} = \frac{(5+5)}{4+4} = \frac{10}{8} = \frac{5}{4} = 1.25$
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