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: 7

Answer

$$\lim_{x\to\infty}h(x)=\lim_{x\to-\infty}h(x)=-\frac{5}{3}$$
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Work Step by Step

$$h(x)=\frac{-5+(7/x)}{3-(1/x^2)}$$ (a) As $x\to\infty$, $x^2$ also approaches $\infty$ and therefore, both $1/x^2$ and $7/x$ will approach $0$. Therefore, $$\lim_{x\to\infty}h(x)=\lim_{x\to\infty}\frac{-5+(7/x)}{3-(1/x^2)}=\frac{-5+0}{3-0}=-\frac{5}{3}$$ (b) As $x\to-\infty$, $x^2$ will approach $\infty$ and therefore, both $1/x^2$ and $7/x$ still approaches $0$. Therefore, $$\lim_{x\to-\infty}h(x)=\lim_{x\to-\infty}\frac{-5+(7/x)}{3-(1/x^2)}=\frac{-5+0}{3-0}=-\frac{5}{3}$$ A graph of the function $h(x)$ is enclosed below, which shows that $h(x)$ approaches $-5/3$ as $x$ approaches either $\infty$ or $-\infty$.
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