Question

What are the asymptote(s) and hole(s), if any, of `f(x) =(1-x)^2/(x^2-1)`?

Answer 1

`f(x)` has a horizontal asymptote `y=1`, a vertical asymptote `x=-1` and a hole at `x=1`.

Explanation:

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

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

with exclusion `x != 1`

As `x->+-oo` the term `2/(x+1) -> 0`, so `f(x)` has a horizontal asymptote `y = 1`.

When `x = -1` the denominator of `f(x)` is zero, but the numerator is non-zero. So `f(x)` has a vertical asymptote `x = -1`.

When `x = 1` both the numerator and denominator of `f(x)` are zero, so `f(x)` is undefined and has a hole at `x=1`. Note that `lim_(x->1) f(x) = 0` is defined. So this is a removable singularity.