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Explain how $x\sin(1/x)$ is not Lipschitz. I know that $\sin(1/x)$ is not Lipschitz and I can show this by proving it is not uniformly continuous, but I'm not sure how to go about things to $x\sin(1/x)$. Thanks!

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    $x\sin\frac1x$ is differentiable on $\mathbb R\setminus\{0\}$. Is its derivative bounded?2012-04-25
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    The derivative is unbounded as x --> 0 so it is not Lipschitz.2012-04-25

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