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

Sep 13, 2015

Find $G \left(x\right) = - \frac{1}{x \left(x + 1\right)}$ with exclusion $x \ne 1$, hence vertical asymptotes $x = - 1$ and $x = 0$ and horizontal asymptote $y = 0$.

#### Explanation:

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

with exclusion $x \ne 1$

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

$G \left(1\right) = \frac{0}{0}$ is undefined. This is a removable singularity - not an asymptote. ${\lim}_{x \to 1} G \left(x\right) = - \frac{1}{2}$ exists.

As $x \to \infty , G \left(x\right) \to 0$, so $y = 0$ is a horizontal asymptote.