Question

How do you identify all of the asymptotes of `G(x) = (x-1)/(x-x^3)`?

Answer 1

Find `G(x) = -1/(x(x+1))` with exclusion `x != 1`, hence vertical asymptotes `x = -1` and `x = 0` and horizontal asymptote `y = 0`.

Explanation:

`G(x) = (x-1)/(x-x^3) = -(x-1)/(x(x-1)(x+1)) = -1/(x(x+1))`

with exclusion `x != 1`

When `x = 0` or `x = -1`, the denominator is zero and the numerator is non-zero, so these are vertical asymptotes.

`G(1) = 0/0` is undefined. This is a removable singularity - not an asymptote. `lim_(x->1) G(x) = -1/2` exists.

As `x->oo, G(x)->0`, so `y = 0` is a horizontal asymptote.