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

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Answer 1 · 2017-12-20T20:04:42

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

Given:

$f(x) = x/(x^4-x^2)$

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

$x/(x^4-x^2) = color(red)(cancel(color(black)(x)))/(color(red)(cancel(color(black)(x))) * x * (x^2-1)) = 1/(x(x-1)(x+1))$

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 $f(x)$ has vertical asymptotes at each of these $x$ values.

As $x->+-oo$ 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]}

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