# What are the asymptote(s) and hole(s), if any, of  f(x) =x/(x^4-x^2)?

Dec 20, 2017

$f \left(x\right)$ has vertical asymptotes $x = - 1$, $x = 0$ and $x = 1$.
It has horizontal asymptote $y = 0$.
It has no slant asymptotes or holes.

#### Explanation:

Given:

$f \left(x\right) = \frac{x}{{x}^{4} - {x}^{2}}$

I like this question, since it provides an example of a rational function which takes a $\frac{0}{0}$ value which is an asymptote rather than a hole...

$\frac{x}{{x}^{4} - {x}^{2}} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{x}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{x}}} \cdot x \cdot \left({x}^{2} - 1\right)} = \frac{1}{x \left(x - 1\right) \left(x + 1\right)}$

Notice that in the simplified form, the denominator is $0$ for $x = - 1$, $x = 0$ and $x = 1$, with the numerator $1$ being non-zero.

So $f \left(x\right)$ has vertical asymptotes at each of these $x$ values.

As $x \to \pm \infty$ the size of the denominator grows without bound, while the numerator stays with $1$. So there is a horizontal asymptote $y = 0$

graph{x/(x^4-x^2) [-10, 10, -5, 5]}