Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Review - Exercises - Page 167: 21

Answer

The asymptotes are: 1) Line $x=0$ is vertical asymptote from the left and $y$ goes to $+\infty$; 2) Line $x=0$ is vertical asymptote from the right and $y$ goes to $+\infty$; 3) Line $y=0$ is horizontal asymptote when $x\to-\infty$; 4) Line $y=0$ is horizontal asymptote when $x\to +\infty$.

Work Step by Step

From the graph below we suspect that the asymptotes are: 1) Line $x=0$ is vertical asymptote from the left and $y$ goes to $+\infty$; 2) Line $x=0$ is vertical asymptote from the right and $y$ goes to $+\infty$; 3) Line $y=0$ is horizontal asymptote when $x\to-\infty$; 4) Line $y=0$ is horizontal asymptote when $x\to +\infty$. Let's prove this: 1) $$\lim_{x\to0^-}\frac{\cos^2x}{x^2}=\left[\frac{\cos^2 0^-}{(0^-)^2}\right]=\left[\frac{1}{0^+}\right]=+\infty$$ which is what is needed. 2) $$\lim_{x\to 0^+}\frac{\cos^2 x}{x^2}=\left[\frac{\cos^20^+}{(0^+)^2}\right]=\left[\frac{1}{0^+}\right]=+\infty$$ which is what is needed. 3) $$\lim_{x\to -\infty}\frac{\cos^2x}{x^2}=\left[\frac{\text{something that is always between $-1$ and $1$}}{(-\infty)^2}\right]=\left[\frac{\text{something that is always between $-1$ and $1$}}{+\infty}\right]=0.$$ which is what is needed. 4) $$\lim_{x\to +\infty}\frac{\cos^2x}{x^2}=\left[\frac{\text{something that is always between $-1$ and $1$}}{(+\infty)^2}\right]=\left[\frac{\text{something that is always between $-1$ and $1$}}{+\infty}\right]=0.$$ which is what is needed.
Small 1511103870
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.