Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 3 - Applications of Differentiation - 3.4 Limits at Infinity; Horizontal Asymptotes - 3.4 Exercises - Page 242: 38

Answer

$x=0, -1 $ and $1$ are vertical asymptotes. $y=-1$ is horizontal asymptotes.

Work Step by Step

\[y=\frac{1+x^4}{x^2-x^4}\] For vertical asymptotes denominator should be zero. \[x^2-x^4=0\] \[\Rightarrow x^2(1-x^2)=0\] \[\Rightarrow x=0,-1 \;\text{and} \;1\] Therefore $x=0, x=-1$ and $x=1$ are vertical asymototes. Let \[L=\lim_{x\rightarrow\infty}\displaystyle\frac{1+x^4}{x^2-x^4}\] \[L=\lim_{x\rightarrow\infty}\frac{\displaystyle\frac{1+x^4}{x^4}}{\displaystyle\frac{x^2-x^4}{x^4}}\] \[L=\lim_{x\rightarrow\infty}\displaystyle\frac{\displaystyle\frac{1}{x^4}+1}{\displaystyle\frac{1}{x^2}-1}\] \[L=\frac{0+1}{0-1}=-1\] Therefore $y=-1$ is horizontal asymptote.
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