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: 41

Answer

$\lim\limits_{x \to -3}\frac{2 - \sqrt {x^{2}-5}}{x+3} = \frac{3}{2} = 1.5$

Work Step by Step

$\frac{2 - \sqrt (x^{2}-5)}{x+3} = \frac{(2 - \sqrt (x^{2}-5))(2 + \sqrt (x^{2}-5))}{(x+3)(2 + \sqrt (x^{2}-5))} = \frac{4-(x^{2}-5)}{(x+3)(2 + \sqrt (x^{2}-5))} = \frac{-(x^{2} - 9)}{(x+3)(2 + \sqrt (x^{2}-5))}$ = $\frac{-(x+3)(x-3)}{(x+3)(2 + \sqrt (x^{2}-5))}$ = $\frac{-(x-3)}{(2 + \sqrt (x^{2}-5))}$ Now, $\lim\limits_{x \to -3}\frac{2 - \sqrt (x^{2}-5)}{x+3} = \lim\limits_{x \to -3}\frac{-(x-3)}{(2 + \sqrt (x^{2}-5))} = \frac{-(-3-3)}{2+2} = \frac{6}{4} = 1.5$
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