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: 38

Answer

$$\frac{1}{4}$$

Work Step by Step

$$\eqalign{ & \mathop {\lim }\limits_{x \to 0} \frac{{ - 1/\left( {x + 2} \right) + 1/2}}{x} \cr & {\text{use the rule }}\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{{\mathop {\lim }\limits_{x \to a} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to a} g\left( x \right)}}\,\,\,\left( {{\text{see page 128}}} \right).{\text{ then}} \cr & \mathop {\lim }\limits_{x \to 0} \frac{{ - 1/\left( {x + 2} \right) + 1/2}}{x} = \frac{{\mathop {\lim }\limits_{x \to 0} \left( { - 1/\left( {x + 2} \right) + 1/2} \right)}}{{\mathop {\lim }\limits_{x \to 0} x}} \cr & {\text{evaluate the limits}}{\text{, replacing }}0{\text{ for }}x \cr & = \frac{{\left( { - 1/\left( {0 + 2} \right) + 1/2} \right)}}{0} = \frac{{ - 1/2 + 1/2}}{0} = \frac{0}{0}{\text{ indeterminate form}} \cr & {\text{Simplifying the numerator as}} \cr & - \frac{1}{{x + 2}} + \frac{1}{2} = \frac{{ - 2 + x + 2}}{{2\left( {x + 2} \right)}} = \frac{x}{{2\left( {x + 2} \right)}} \cr & then \cr & \mathop {\lim }\limits_{x \to 0} \frac{{ - 1/\left( {x + 2} \right) + 1/2}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{x}{{2\left( {x + 2} \right)}}}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{x}{{2x\left( {x + 2} \right)}} \cr & = \mathop {\lim }\limits_{x \to 0} \frac{1}{{2\left( {x + 2} \right)}} \cr & {\text{evaluating the limit when }}x \to 0 \cr & = \frac{1}{{2\left( {0 + 2} \right)}} \cr & = \frac{1}{4} \cr} $$
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