Calculus with Applications (10th Edition)

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

Chapter 3 - The Derivative - 3.1 Limits - 3.1 Exercises - Page 137: 37

Answer

$$ - \frac{1}{9}$$

Work Step by Step

$$\eqalign{ & \mathop {\lim }\limits_{x \to 0} \frac{{1/\left( {x + 3} \right) - 1/3}}{x} \cr & {\text{use the rule }}\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{{\mathop {\lim }\limits_{x \to a} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to a} g\left( x \right)}}\,\,\,\left( {{\text{see page 128}}} \right).{\text{ then}} \cr & \mathop {\lim }\limits_{x \to 0} \frac{{1/\left( {x + 3} \right) - 1/3}}{x} = \frac{{\mathop {\lim }\limits_{x \to 0} \left( {1/\left( {x + 3} \right) - 1/3} \right)}}{{\mathop {\lim }\limits_{x \to 0} x}} \cr & {\text{evaluate the limits}}{\text{, replacing }}0{\text{ for }}x \cr & = \frac{{\left( {1/\left( {0 + 3} \right) - 1/3} \right)}}{0} = \frac{{1/3 - 1/3}}{0} = \frac{0}{0}{\text{ indeterminate form}} \cr & {\text{Simplifying the numerator as}} \cr & \frac{1}{{x + 3}} - \frac{1}{3} = \frac{{3 - \left( {x + 3} \right)}}{{3\left( {x + 3} \right)}} = \frac{{ - x}}{{3\left( {x + 3} \right)}} \cr & then \cr & \mathop {\lim }\limits_{x \to 0} \frac{{1/\left( {x + 3} \right) - 1/3}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{ - x}}{{3\left( {x + 3} \right)}}}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{ - x}}{{3x\left( {x + 3} \right)}} \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{ - 1}}{{3\left( {x + 3} \right)}} \cr & {\text{evaluating the limit when }}x \to 0 \cr & = \frac{{ - 1}}{{3\left( {0 + 3} \right)}} \cr & = - \frac{1}{9} \cr} $$
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