How do you find the asymptotes for #f(x)=x/(x^2+4)#?

1 Answer
Sep 16, 2016

Answer:

There is no vertical asymptote, as the denominator will always be unequal to #0# (even #f(x)>=4#, whatever the value of #x#).

Explanation:

For the horizontal asymptote, we look what happens if #x# grows very large (both positive and negative).

The #+4# in the denominator will make less and less of a difference, and the function will look more and more like:
#f(x)=x/x^2=1/x#

In "the language":

#lim_(x->oo) f(x)=0# and #lim_(x->-oo) f(x)=0#

So there's only #y=0# as horizontal asymptote at the #+-oo# ends of the function.
For #x=0->f(x)=0#, so it would seem to be not a proper asymptote (in the meaning "never there but as close as you want"), but it still counts as one, because for the rest of the function it really is "never there".
graph{x/(x^2+4) [-10, 10, -5, 5]}