Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 1 - Functions and Limits - 1.6 Calculating Limits Using the Limit Laws - 1.6 Exercises - Page 70: 27


$\lim\limits_{x \to 16}\frac{4-\sqrt x}{16x-x^{2}}=\frac{1}{128}$

Work Step by Step

$\lim\limits_{x \to 16}\frac{4-\sqrt x}{16x-x^{2}}=\lim\limits_{x \to 16}\frac{4-\sqrt x}{x(16-x)}=\lim\limits_{x \to 16}\frac{4-\sqrt x}{x(4-\sqrt x)(4+\sqrt x)}=\lim\limits_{x \to 16}\frac{1}{x(4+\sqrt x)}=\frac{1}{16\times(4+4)}=\frac{1}{128}$
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