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 - Practice Exercises - Page 101: 47


(a) $ x= 3$ is a vertical asymptote (b) $ x= 1$ is a vertical asymptote (c) $ x=-4$ is a vertical asymptote

Work Step by Step

(a) Since $ y=\frac{x^{2}+4}{x-3}$ is undefined at $ x=3$ : $ \lim _{x \rightarrow 3^{-}} \frac{x^{2}+4}{x-3}=-\infty $ and $\lim _{x \rightarrow 3^{+}} \frac{x^{2}+4}{x-3}=+\infty$ Thus, $ x=3$ is a vertical asymptote. (b) $ y=\frac{x^{2}-x-2}{x^{2}-2 x+1}$ is undefined at $ x=1 $: $ \lim _{x \rightarrow 1} \frac{x^{2}-x-2}{x^{2}-2 x+1}=-\infty $ and $\lim _{x \rightarrow 1^{+}} \frac{x^{2}-x-2}{x^{2}-2 x+1}=-\infty$ Thus, $ x=1$ is a vertical asymptote. (c) Since $ y=\frac{x^{2}+x-6}{x^{2}+x-8}$ is undefined at $ x=2$ and $-4$: $\begin{align*}\lim _{x \rightarrow 2} \frac{x^{2}+x-6}{x^{2}+2 x-8}&=\lim _{x \rightarrow 2} \frac{x+3}{x+4}=\frac{5}{6}\\ \lim _{x \rightarrow-4} \frac{x^{2}+x-6}{x^{2}+2 x-8}&=\lim _{x \rightarrow-4^{-}} \frac{x+3}{x+4}=\infty\end{align*}$ $\lim _{x \rightarrow-4^{+}} \frac{x^{2}+x-6}{x^{2}+2 x-8}=\lim _{x \rightarrow-4^{+}} \frac{x+3}{x+4}=-\infty .$ Thus, $ x=-4$ is a vertical asymptote.
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