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 109: 78


The graph of $f(x)/g(x)$ can still have a horizontal or oblique asymptote, just not the vertical one.

Work Step by Step

The graph of $f(x)/g(x)$ cannot give out any vertical asymptote if $g(x)$ is never $0$, but it is still possible that the graph has other types of asymptote (horizontal asymptote for example). - The graph of $f(x)/g(x)$ does not have vertical asymptote if $g(x)$ is never $0$, because only by having values of $x=a$ that makes $g(x)=0$ could we have as $x\to a$, $g(x)\to0$, making $f(x)/g(x)$ approach $\pm\infty$, and the graph having a vertical asymptote. - However, other types of asymptote are still possible. I would give a function of this type that has a horizontal asymptote: $$\frac{f(x)}{g(x)}=\frac{x^2}{x^2+1}$$ Here because $x^2\gt0$ for all $x$, $x^2+1$ is never $0$. $$\lim_{x\to\pm\infty}\frac{f(x)}{g(x)}=\lim_{x\to\pm\infty}\frac{x^2}{x^2+1}$$ Divide both numerator and denominator by $x^2$: $$\lim_{x\to\pm\infty}\frac{f(x)}{g(x)}=\lim_{x\to\pm\infty}\frac{1}{1+\frac{1}{x^2}}=\frac{1}{1+0}=1$$ So $y=1$ is the horizontal asymptote of $f(x)/g(x)$. In fact, if the highest degree of $x$ of $f(x)$ is smaller or equal to that of $g(x)$, we would always have the horizontal asymptote of $f(x)/g(x)$.
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