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: 49

Answer

$$\infty $$

Work Step by Step

$$\eqalign{ & \mathop {\lim }\limits_{x \to \infty } \frac{{2{x^3} - x - 3}}{{6{x^2} - x - 1}} \cr & {\text{evaluating the limit use the rules 4 and 2 }}\left( {page\,\,\,128} \right) \cr & = \frac{{\mathop {\lim }\limits_{x \to \infty } 2{x^3} - \mathop {\lim }\limits_{x \to \infty } x - \mathop {\lim }\limits_{x \to \infty } 3}}{{\mathop {\lim }\limits_{x \to \infty } 6{x^2} - \mathop {\lim }\limits_{x \to \infty } x - \mathop {\lim }\limits_{x \to \infty } 1}} \cr & = \frac{{2{{\left( \infty \right)}^3} - \infty - 3\left( \infty \right)}}{{6{{\left( \infty \right)}^2} - \infty - 1}} \cr & = \frac{\infty }{\infty } \cr & {\text{divive the numerator and denominator by the highest power of }}x{\text{ in}} \cr & {\text{the denominator}}{\text{. the highhest power is }}{x^2} \cr & = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{2{x^3}}}{{{x^2}}} - \frac{x}{{{x^2}}} - \frac{3}{{{x^2}}}}}{{\frac{{6{x^2}}}{{{x^2}}} - \frac{x}{{{x^2}}} - \frac{1}{{{x^2}}}}} \cr & = \mathop {\lim }\limits_{x \to \infty } \frac{{2x - \frac{1}{x} - \frac{3}{{{x^2}}}}}{{6 - \frac{1}{x} - \frac{1}{{{x^2}}}}} \cr & {\text{use the rule }}\mathop {\lim }\limits_{x \to \infty } \frac{1}{{{x^n}}} = 0 \cr & {\text{the denominator approaches to 6}}{\text{, while the numerator becomes a }} \cr & {\text{positive number that is larger and larger in magnitude}}{\text{, so}} \cr & \mathop {\lim }\limits_{x \to \infty } \frac{{2{x^3} - x - 3}}{{6{x^2} - x - 1}} = \infty \cr} $$
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