## Thomas' Calculus 13th Edition

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