# What are the asymptotes of g(x) = (x-1)/(x-x^3)?

Jun 16, 2018

The vertical asymptotes are -1, 0 and 1, The horizontal asymptote is y=0.

#### Explanation:

Vertical asymptotes:
$x - {x}^{3} = 0$ then $x \left(1 - {x}^{2}\right) = 0$ and $x \left(1 - x\right) \left(1 + x\right) = 0$
Are: ${x}_{1} = 0$; ${x}_{2} = 1$ and ${x}_{3} = - 1$

Horizontal asymptote:
$L i {m}_{x \setminus \rightarrow + \setminus \infty} \setminus \frac{x - 1}{x - {x}^{3}} = \setminus \frac{+ \setminus \infty}{- \setminus \infty}$ IND
$L i {m}_{x \setminus \rightarrow + \setminus \infty} \setminus \frac{x - 1}{x \left(1 - x\right) \left(1 + x\right)} = L i {m}_{x \setminus \rightarrow + \setminus \infty} \setminus \frac{1}{x \left(1 + x\right)} = \setminus \frac{1}{+ \setminus \infty} = 0$

$L i {m}_{x \setminus \rightarrow - \setminus \infty} \setminus \frac{x - 1}{x - {x}^{3}} = \setminus \frac{- \setminus \infty}{\setminus \infty}$ IND
$L i {m}_{x \setminus \rightarrow - \setminus \infty} \setminus \frac{x - 1}{x \left(1 - x\right) \left(1 + x\right)} = L i {m}_{x \setminus \rightarrow - \setminus \infty} \setminus \frac{1}{x \left(1 + x\right)} = \setminus \frac{1}{- \setminus \infty} = 0$