Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 3 - The Derivative - Chapter Review - Review Exercises - Page 189: 30

Answer

$$\frac{1}{8}$$

Work Step by Step

$$\eqalign{ & \mathop {\lim }\limits_{x \to 16} \frac{{\sqrt x - 4}}{{x - 16}} \cr & {\text{use the rule }}\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{{\mathop {\lim }\limits_{x \to a} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to a} g\left( x \right)}}\,\,\,\left( {{\text{see page 128}}} \right).{\text{ Then}} \cr & \mathop {\lim }\limits_{x \to 16} \frac{{\sqrt x - 4}}{{x - 16}} = \frac{{\mathop {\lim }\limits_{x \to 16} \left( {\sqrt x - 4} \right)}}{{\mathop {\lim }\limits_{x \to 16} \left( {x - 16} \right)}} \cr & {\text{evaluate the limits}}{\text{, replacing }}16{\text{ for }}x \cr & \mathop {\lim }\limits_{x \to 16} \frac{{\sqrt x - 4}}{{x - 16}} = \frac{{\sqrt {16} - 4}}{{16 - 16}} = \frac{0}{0}{\text{ indeterminate form}} \cr & {\text{rationalizing}} \cr & = \mathop {\lim }\limits_{x \to 16} \frac{{\sqrt x - 4}}{{x - 16}} \times \frac{{\sqrt x + 4}}{{\sqrt x + 4}} \cr & = \mathop {\lim }\limits_{x \to 16} \frac{{\left( {\sqrt x - 4} \right)\left( {\sqrt x + 4} \right)}}{{\left( {x - 16} \right)\left( {\sqrt x + 4} \right)}} \cr & = \mathop {\lim }\limits_{x \to 16} \frac{{{{\left( {\sqrt x } \right)}^2} - {{\left( 4 \right)}^2}}}{{\left( {x - 16} \right)\left( {\sqrt x + 4} \right)}} \cr & = \mathop {\lim }\limits_{x \to 16} \frac{{x - 16}}{{\left( {x - 16} \right)\left( {\sqrt x + 4} \right)}} \cr & = \mathop {\lim }\limits_{x \to 16} \frac{1}{{\sqrt x + 4}} \cr & {\text{evaluating the limit when }}x \to 9 \cr & = \frac{1}{{\sqrt {16} + 4}} \cr & = \frac{1}{8} \cr} $$
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