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]}