Answer
The function doesn't have a horizontal or oblique asymptote.
$x=0$ is a vertical asymptote
Work Step by Step
$R(x)=\frac{x^4-1}{x^2-x},$
The degree of the leading coefficient of the numerator is, $n=4$. the degree of the leading coefficient the denominator is, $m=2$.
Thus, $n\geq m+2,$
Therefore, The function doesn't have a horizontal or oblique asymptote.
We have:
$\frac{(x-1)(x+1)(x^2+1)}{x(x-1)}=\frac{(x+1)(x^2+1)}{x}$
Thus, $x=0$ is a vertical asymptote.
