Question

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

Answer 1

Since `f(x) = ((x-1)(x+1))/(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=+-1`

Explanation:

Except for the values `x=1` and `x=-1` where `f(x)` becomes equivalent to `0/0`

`f(x)=1` (since `(x-1)(x+1) = (x^2-1)`)