Use this sequence: .
Yeah that makes sense. I don't know what I was thinking. I think in my head I wanted to do a proof as you would when you are proving that a function is continuous as opposed to discontinuous. But just to clarify my last question, in general it seems that any sequence of type (1/n^p) where p>0 or a^n where |a| < 1 are likely candidates for a contradiction.
Here's my complete answer by the way:
f(x0) = f(0) = 0
(xn) = 1/n^2 subset of dom(f) = R
(xn) -> 0 = x0 (can be easily proven using defenition of a limit)
lim f(xn) = lim 1/squareroot(1/n^2) = +infinity does not equal f(x0) = f(0) = 0.
You are basically looking for a sequence that goes to 0 but f(xn) goes to something other than 0 to prove discontinuity.
Thanks that was very helpful. I've got another question this time on proving that a function is continuous. Your help was much appreciated. Thanks