TITLE: Determine whether or not the cube root of x is Lipschitz.
QUESTION [5 upvotes]: Determine whether or not the cube root of x is Lipschitz.   x^(1/3)

I understand that this is NOT Lipschitz. I am having trouble properly proving this. I know for other problems I have shown functions are not uniformly continuous to prove that they are not Lipschitz, but I can't seem to figure out the right way to go about this problem. Reviewing for a test, Thanks!

REPLY [3 votes]: Too see it without using derivatives, observe that
$$ x^{1/3}-y^{1/3} = (x-y)/(x^{2/3} + (xy)^{1/3} + y^{2/3} ) $$
Now as x or y goes to zero, you can make the denominator as large as you wish (as cube root is a continuous function at zero). Thus, no single L will suffice.