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: 33

Answer

$$\frac{3}{8}$$

Work Step by Step

$$\eqalign{ & \mathop {\lim }\limits_{x \to - \infty } \left( {\frac{3}{8} + \frac{3}{x} - \frac{6}{{{x^2}}}} \right) \cr & {\text{use the sum rule for limits }}\mathop {\lim }\limits_{x \to a} \left[ {f\left( x \right) \pm g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to a} f\left( x \right) \pm \mathop {\lim }\limits_{x \to a} g\left( x \right) \cr & = \mathop {\lim }\limits_{x \to - \infty } \frac{3}{8} + \mathop {\lim }\limits_{x \to - \infty } \frac{3}{x} - \mathop {\lim }\limits_{x \to - \infty } \frac{6}{{{x^2}}} \cr & = \mathop {\lim }\limits_{x \to - \infty } \frac{3}{8} + 3\mathop {\lim }\limits_{x \to - \infty } \frac{1}{x} - 6\mathop {\lim }\limits_{x \to - \infty } \frac{1}{{{x^2}}} \cr & {\text{evaluate the limits}}{\text{, use the rules }}\mathop {\lim }\limits_{x \to \infty } k = k{\text{ and }}\mathop {\lim }\limits_{x \to - \infty } \frac{1}{{{x^n}}} = 0 \cr & = \frac{3}{8} + 3\left( 0 \right) - 6\left( 0 \right) \cr & = \frac{3}{8} \cr} $$
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