# How do you find vertical, horizontal and oblique asymptotes for (x-1)/(x-x^3)?

Nov 17, 2016

The vertical asymptotes are $x = 0$ and $x = - 1$

The horizontal asymptote is $y = 0$

#### Explanation:

Let's simplify the expression

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

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

So, we have a hole at $\left(x = 1\right)$

A we cannot divide by $0$,

$x \ne 0$ and $x \ne - 1$

We have two vertical asymptotes $x = 0$ and $x = - 1$

As the degree of the numerator $<$ degree of the denominator,
we don't have oblique asymptotes.

${\lim}_{x \to \pm \infty} - \frac{1}{x} ^ 2 = {0}^{-}$

The horizontal asymptote is $y = 0$
graph{-1/(x(1+x)) [-25.66, 25.65, -12.83, 12.84]}