## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 2 - Section 2.6 - Limits Involving Infinity; Asymptotes of Graphs - Exercises - Page 109: 103

#### Answer

- Vertical asymptote: $x=0$ - Oblique asymptote: $y=x$ #### Work Step by Step

$$y=\frac{x^2-1}{x}$$ - As $x\to0$, $x^2-1$ approaches $(0)^2-1=-1\lt0$ , so $(x^2-1)/x$ will approach $-\infty$. In other words, $$\lim_{x\to0}\frac{x^2-1}{x}=-\infty$$ 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: $$y=\frac{(x^2)-1}{x}$$ $$y=\frac{x^2}{x}-\frac{1}{x}$$ $$y=x-\frac{1}{x}$$ As $x\to\pm\infty$, $1/x$ 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. 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.