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: 44

Answer

$$2$$

Work Step by Step

$$\eqalign{ & \mathop {\lim }\limits_{x \to - \infty } \frac{{8x + 2}}{{4x - 5}} \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 - \infty } \frac{{8x + 2}}{{4x - 5}} = \frac{{\mathop {\lim }\limits_{x \to - \infty } \left( {8x + 2} \right)}}{{\mathop {\lim }\limits_{x \to - \infty } \left( {4x - 5} \right)}} \cr & {\text{evaluate the limits}}{\text{, replacing }}\infty {\text{ for }}x \cr & \frac{{\mathop {\lim }\limits_{x \to - \infty } \left( {8x + 2} \right)}}{{\mathop {\lim }\limits_{x \to - \infty } \left( {4x - 5} \right)}} = \frac{{8\left( { - \infty } \right) + 2}}{{4\left( { - \infty } \right) - 1}} = \frac{{ - \infty }}{{ - \infty }}{\text{ indetermiante form}} \cr & {\text{divide each term in the numerator and denominator by }}x \cr & = \mathop {\lim }\limits_{x \to - \infty } \frac{{\frac{{8x}}{x} + \frac{2}{x}}}{{\frac{{4x}}{x} - \frac{5}{x}}} \cr & = \mathop {\lim }\limits_{x \to - \infty } \frac{{8 + \frac{2}{x}}}{{4 - \frac{5}{x}}} = \frac{{\mathop {\lim }\limits_{x \to - \infty } 8 + \mathop {\lim }\limits_{x \to - \infty } \frac{2}{x}}}{{\mathop {\lim }\limits_{x \to - \infty } 4 - \mathop {\lim }\limits_{x \to - \infty } \frac{5}{x}}} \cr & = \frac{{\mathop {\lim }\limits_{x \to - \infty } 8 + 2\mathop {\lim }\limits_{x \to - \infty } \frac{1}{x}}}{{\mathop {\lim }\limits_{x \to - \infty } 4 - 5\mathop {\lim }\limits_{x \to - \infty } \frac{1}{x}}} \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{{\mathop {\lim }\limits_{x \to - \infty } 8 + 2\mathop {\lim }\limits_{x \to - \infty } \frac{1}{x}}}{{\mathop {\lim }\limits_{x \to - \infty } 4 - 5\mathop {\lim }\limits_{x \to - \infty } \frac{1}{x}}} = \frac{{8 + 2\left( 0 \right)}}{{4 - 5\left( 0 \right)}} = 2 \cr} $$
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