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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- Nov 2nd 2009, 06:38 PMdannyboycurtislipschitz functions
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. - Nov 2nd 2009, 08:29 PMdannyboycurtis
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 - Nov 3rd 2009, 04:36 AMchisigma
Necessary condition for a function to be 'Lipschitz' is that its derivative is

__bounded__in all the interval of definition...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$