## Calculus: Early Transcendentals 8th Edition

There are no horizontal asymptotes. $x=5~$ is a vertical asymptote.
Horizontal asymptotes: $\lim\limits_{x \to -\infty}\frac{x^3-x}{x^2-6x+5} = -\infty$ $\lim\limits_{x \to \infty}\frac{x^3-x}{x^2-6x+5} = \infty$ There are no horizontal asymptotes. Vertical asymptotes: $\frac{x^3-x}{x^2-6x+5} = \frac{(x)(x-1)(x+1)}{(x-5)(x-1)}$ $x=5~~$ is a vertical asymptote. Although $(x-1)$ is in the denominator, we can see that $(x-1)$ is also in the numerator. Therefore, $~~x=1~~$ is not a vertical asymptote, even though the graph is undefined at this point.