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: 104

Answer

- Vertical asymptote: $x=0$ - Oblique asymptote: $y=x$

Work Step by Step

$$y=\frac{x^3+1}{x^2}$$ - As $x\to0$, $x^2$ approaches $0$ and $x^3+1$ approaches $(0)^3+1=1\gt0$ , so $(x^3+1)/x^2$ will approach $\infty$. In other words, $$\lim_{x\to0}\frac{x^3+1}{x^2}=\infty$$ Therefore, the line $x=0$ is the vertical asymptote of the graph. - There is one more oblique asymptote, but to find it, we need to change the form of the function: $$y=\frac{(x^3)+1}{x^2}$$ $$y=\frac{x^3}{x^2}+\frac{1}{x^2}$$ $$y=x+\frac{1}{x^2}$$ As $x\to\pm\infty$, $1/x^2$ approaches $0$, making the function turn into $$y=x$$ which is the oblique asymptote of the graph. The graph, along with 2 asymptotes, is shown below.
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.