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

Published by Pearson

# Chapter 2 - Section 2.6 - Limits Involving Infinity; Asymptotes of Graphs - Exercises - Page 109: 100

#### Answer

- Vertical asymptote: $x=1$ - Oblique asymptote: $y=x+1$ #### Work Step by Step

$$y=\frac{x^2+1}{x-1}$$ - As $x\to1$, $(x-1)$ approaches $0$ and $x^2+1$ approaches $2\gt0$ , so $(x^2+1)/(x-1)$ will approach $\infty$. In other words, $$\lim_{x\to1}\frac{x^2+1}{x-1}=\infty$$ Therefore, the line $x=1$ 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: $$y=\frac{(x^2-1)+2}{x-1}$$ $$y=\frac{x^2-1}{x-1}+\frac{2}{x-1}$$ $$y=(x+1)+\frac{2}{x-1}$$ As $x\to\pm\infty$, $2/(x-1)$ approaches $0$, making the function turn into $$y=x+1$$ which is the oblique asymptote of the graph. The graph, along with 2 asymptotes, are 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.