# Does f(x)=((x-1)(x+1))/(x^2-1) have an asymptote or a hole?

Since $f \left(x\right) = \frac{\left(x - 1\right) \left(x + 1\right)}{{x}^{2} - 1}$ is not a curve, it does not have an asymptote (using most common definitions of the term.
It is undefined (has [in this case "removable"] holes) for $x = \pm 1$
Except for the values $x = 1$ and $x = - 1$ where $f \left(x\right)$ becomes equivalent to $\frac{0}{0}$
$f \left(x\right) = 1$ (since $\left(x - 1\right) \left(x + 1\right) = \left({x}^{2} - 1\right)$)