Question

How do you find the asymptote(s) or holes of `f(x) = (x-1) / (x^2-1)`?

Answer 1

We have a vertical asymptote at `x+1=0` and no horizontal or slanting asymptote. We have a hole at `x=1`.

Explanation:

`(x-1)/(x^2-1)=((x-1))/((x-1)(x+1))=1/(x+1)`

Although `1/(x+1)` may be defined for `x=1`, as `x-1` has cancelled out, the fact is for `x=1`, we do not have `(x-1)/(x^2-1)`. Hence, we have a hole a `x=1`.

As we have `(x+1)` in denominator, we have a vertical asymptote `x+1=0`.

As the degree of numerator is less than that of denominator, we do not have any horizontal or slanting asymptote.

graph{(x-1)/(x^2-1) [-10, 10, -5, 5]}