What are the asymptotes for #f(x)= x/(x(x-2))#?

1 Answer
Jul 13, 2015

Answer:

#f(x) = x/(x(x-2)) = 1/(x-2)# with exclusion #x != 0#.

So there's one vertical asymptote (simple pole) at #x = 2#

The horizontal asymptote is #y = 0#.

Explanation:

For any rational expression with polynomial numerator and denominator, there are usually vertical asymptotes whenever the denominator is zero.

Here we see an exception to that in that both the numerator and denominator of #f(x)# are zero when #x=0#. So #f(0) = 0/0# is undefined. This is a removable singularity in that we can redefine #f(0) = -1/2# to make #f# well defined and continuous at #x=0#.

Since #f(x) = 1/(x-2)# (except for #x = 0#),

we can see that #f(x)->0# as #x->+-oo#,

resulting in a horizontal asymptote #y = 0#.