I need to show that $\displaystyle f(x)=\sqrt[3]{x}$ and $\displaystyle g(x)=xsin(\frac{1}{x})$ for $\displaystyle x\in (0,1]$ are NOT Lipschitz. Im a little lost on how to proceed.
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is it enough to show that they are not uniformly continuous? if so Im still not sure how to show that f(x) is not uniformly continous
Necessary condition for a function to be 'Lipschitz' is that its derivative is bounded in all the interval of definition... Kind regards $\displaystyle \chi$ $\displaystyle \sigma$
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