I don't agree that this discussion is tedious. I think your reply was rather tedious. Personally I don't beleive that $\displaystyle 0^{0} := 1 $, I think that this definition only holds for convenience in special cases.

It's not hard to think of examples where $\displaystyle f(x), g(x) $ approach zero continuously, yet $\displaystyle f(x)^{g(x)} $ does not approach $\displaystyle 1$

I'm still interested in your rigorous proof though.

-pomp.