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: 40

Answer

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

Work Step by Step

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