# How do you find the asymptotes for G(x) = (x-1)/(x-x^3)?

Nov 23, 2016

There is a hole at $x = - 1$
The vertical asymptotes are $x = 0$ and $x = 1$
No oblique asymptote.
The horizontal asymptote is $y = 0$

#### Explanation:

Let's factorise the numerator and denominator

$x - 1 = - \left(1 + x\right)$

$x - {x}^{3} = x \left(1 - {x}^{2}\right) = x \left(1 - x\right) \left(1 + x\right)$

So,

$G \left(x\right) = \frac{x - 1}{x \left(1 - {x}^{2}\right)} = - \frac{\cancel{1 + x}}{x \left(1 - x\right) \cancel{1 + x}}$

$= - \frac{1}{x \left(1 - x\right)}$

There is a hole at $x = - 1$

As we cannot divide by $0$,
so, $x \ne 0$ and $x \ne 1$

Therefore, $x = 0$ and $x = 1$ are vertical asymptotes

As the degree of the numerator is $>$ the degree of the denominator, there is no slant asymptote.

For the limits, we take the terms of highest degree-

${\lim}_{x \to \pm \infty} G \left(x\right) = {\lim}_{x \to \pm \infty} - \frac{1}{x} ^ 2 = {0}^{-}$

Therefore, $y = 0$ is a horizontal asymptote

graph{(x-1)/(x(1-x^2)) [-18.16, 13.88, -7.75, 8.27]}