University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Section 2.6 - Limits Involving Infinity; Asymptotes of Graphs - Exercises - Page 107: 6



Work Step by Step

$$g(x)=\frac{1}{8-\frac{5}{x^2}}$$ (a) As $x\to\infty$, $x^2$ also approaches $\infty$ and $5/x^2$ will approach $0$. Therefore, $$\lim_{x\to\infty}g(x)=\lim_{x\to\infty}\frac{1}{8-\frac{5}{x^2}}=\frac{1}{8-0}=\frac{1}{8}$$ (b) As $x\to-\infty$, $x^2$ will approach $\infty$ and $5/x^2$ approaches $0$ as a result. Therefore, $$\lim_{x\to-\infty}g(x)=\lim_{x\to-\infty}\frac{1}{8-\frac{5}{x^2}}=\frac{1}{8-0}=\frac{1}{8}$$ A graph of the function $g(x)$ is enclosed below, which shows that $g(x)$ approaches $1/8$ as $x$ approaches either $\infty$ or $-\infty$.
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