Question

What are the asymptote(s) and hole(s), if any, of `f(x) =(x-1)/(x^4-1)`?

Answer 1

`f(x)` has a vertical asymptote at `x=-1`, a hole at `x=1` and a horizontal asymptote `y=0`. It has no oblique asymptotes.

Explanation:

`f(x) = (x-1)/(x^4-1)`

`color(white)(f(x)) = color(red)(cancel(color(black)((x-1))))/(color(red)(cancel(color(black)((x-1))))(x+1)(x^2+1))`

`color(white)(f(x)) = 1/((x+1)(x^2+1))`

with exclusion `x!=-1`

Note that `x^2+1 > 0` for any real values of `x`

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

When `x=1` both the numerator and denominator of the defining expression for `f(x)` are zero, but the simplified expression is well defined and continuous at `x=1`. So there is a hole at `x=1`.

As `x->+-oo` the denominator of the simplified expression `->oo`, while the numerator is constant `1`. Hence the function tends to `0` and has a horizontal asymptote `y=0`

`f(x)` has no oblique (a.k.a. slant) asymptotes. In order for a rational function to have an oblique asymptote, the numerator must have degree exactly one more than the denominator.

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