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

Published by Pearson

# 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.

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