Answer
There are no horizontal asymptotes.
$x=5~$ is a vertical asymptote.
Work Step by Step
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.
