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!
Explain how $x\sin(1/x)$ is not Lipschitz.
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real-analysis
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3$x\sin\frac1x$ is differentiable on $\mathbb R\setminus\{0\}$. Is its derivative bounded? – 2012-04-25
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3The derivative is unbounded as x --> 0 so it is not Lipschitz. – 2012-04-25