Are you looking for a rigorous epsilon-delta proof, or just a round-about proof? (for example, you can just show that the derivative is unbounded)
I suppose I was mixing up my definitions of uniformly continuous and Lipschitz. Sorry for the confusion.
on [0,1] is uniformly continuous, but not Lipschitz.
Nevermind... making too many mistakes, will think through before I post again hahah.
HallsofIvy: I liked your solution, although I don't get the 'picture' of the solution:
I need to find an Epsylon for which for every delta>0 :
|x1-x2|<delta ----> |f(x1)-f(x2)|>=Epsilon
now, if I create two sequences:
How can I move on, in order to show that there is an Epsilon for which :