Answer
$x=1, x=5$ are vertical asymptotes.
There is no horizontal asymptotes.
Work Step by Step
\[y=\frac{x^3-x}{x^2-6x+5}\]
For vertical asymptotes denominator should be zero.
\[x^2-6x+5=0\]
\[\Rightarrow x^2-5x-x+5=0\]
\[\Rightarrow (x-5)(x-1)=0\]
\[\Rightarrow x=1 \;\text{and} \;5\]
Therefore $x=1, x=5$ are vertical asymototes.
Let \[L=\lim_{x\rightarrow\infty}\displaystyle\frac{x^3-x}{x^2-6x+5}\]
\[L=\lim_{x\rightarrow\infty}\frac{\displaystyle\frac{x^3-x}{x^2}}{\displaystyle\frac{x^2-6x+5}{x^2}}\]
\[L=\lim_{x\rightarrow\infty}\displaystyle\frac{\displaystyle x-\frac{1}{x}}{1-\displaystyle\frac{6}{x}+\displaystyle\frac{5}{x^2}}\]
\[L=\infty\]
Which is not finite therefore there is no horizontal asymptotes.
