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: 35

Answer

$\lim\limits_{x \to 9}\frac{\sqrt x - 3}{x-9} = \frac{1}{6} = 0.16$

Work Step by Step

$\frac{\sqrt x - 3}{x-9} = \frac{\sqrt x - 3}{(\sqrt x)^{2}-3^{2}} = \frac{\sqrt x - 3}{(\sqrt x-3)(\sqrt x+3)} = \frac{1}{(\sqrt x+3)}$ Now, $\lim\limits_{x \to 9}\frac{\sqrt x - 3}{x-9} = \frac{1}{(\sqrt x+3)} = \frac{1}{3+3} = \frac{1}{6} = 0.16$
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