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: 39

Answer

$\lim\limits_{x \to 2}\frac{\sqrt (x^{2}+12) -4}{x-2} =\frac{1}{2} = 0.5$

Work Step by Step

$\frac{\sqrt (x^{2}+12) -4}{x-2} = \frac{(\sqrt (x^{2}+12) -4)(\sqrt (x^{2}+12) +4)}{(x-2)(\sqrt (x^{2}+12) +4)} = \frac{x^{2}+12-16}{(x-2)(\sqrt (x^{2}+12) +4)} = \frac{x^{2}-4}{(x-2)(\sqrt (x^{2}+12) +4)}$ = $\frac{(x+2)(x-2)}{(x-2)(\sqrt (x^{2}+12) +4)}$ = $\frac{(x+2)}{(\sqrt (x^{2}+12) +4)}$ Now, $\lim\limits_{x \to 2}\frac{\sqrt (x^{2}+12) -4}{x-2} = \lim\limits_{x \to 2}\frac{(x+2)}{(\sqrt (x^{2}+12) +4)} = \frac{2+2}{4+4} = \frac{4}{8} = \frac{1}{2} = 0.5$
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