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

Answer 1

#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#.

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#.

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Answer 2

James Allen

The function (f(x) = \frac{x}{x(x-2)}) has two vertical asymptotes at (x = 0) and (x = 2). There are no horizontal or slant asymptotes for this function.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.