Following function is given: $$f(x) = \begin{cases} x \sin\left(\frac{1}{x}\right) & \text{if $x\ne 0$, and} \\ 0 & \text{if $x=0$.}\end{cases}$$ Now we have to show, that this function is Lipschitz-continuous at $x = 0$.
Following was our try, please let us know if there are any mistakes:
Let $x\ne 0$ and $y=0$. Then $$|f(y)-f(x)|=\left|x\sin\left(\frac{1}{x}\right)\right| \le |x| = |x-y|=1\cdot|x-y|$$
Therefore $L=1$ should be Lipschitz-constant and the function is locally continuous at $y=0$.
Is this right? Otherwise, we would be very happy about getting more useful hints.