Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 2: Limits and Continuity - Section 2.6 - Limits Involving Infinity; Asymptotes of Graphs - Exercises 2.6 - Page 98: 64

Answer

Graph:

Work Step by Step

The degree of the numerator is 0, which is less than the degree of the denominator (1), so we can write $y=f(x)=0+\displaystyle \frac{1}{x-1}$ $\displaystyle \lim_{x\rightarrow+\infty}f(x)=0, \quad \lim_{x\rightarrow-\infty}f(x)=0,$ so, the horizontal asymptote is $y=0.$ f(x) is not defined for $x=-1$. Vertical asymptote : $x=-1$, with $\displaystyle \lim_{x\rightarrow-1^{-}}f(x)=-\infty, \quad \displaystyle \lim_{x\rightarrow-1^{+}}f(x)=+\infty$ Plot some points: y-intercept: $f(0)=1,$ plot ($0,1)$. Also, use $f(-2)=1$ to plot $(-2,1).$ Graph the asymptotes, and use the behavior in the vicinity of the asymptotes to sketch the graph.
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