Originally Posted by

**Laurent** I don't think this is true... (the derivative of $\displaystyle x\mapsto x\sin\frac{1}{x}$ gets arbitrarily large on neighbourhoods of $\displaystyle 0$)

On the other hand, you can say: the function $\displaystyle f:x\mapsto f(x)=x\sin\frac{1}{x}$ is continuous on $\displaystyle (0,1]$, and $\displaystyle f(x)\to_{x\to 0^+} 0$ (because $\displaystyle |f(x)|\leq x$), so that $\displaystyle f$ can be extended to a continuous function on $\displaystyle [0,1]$ by letting $\displaystyle f(0)=0$. Now, we have a continuous function on the compact subset $\displaystyle [0,1]$, so it is uniformly continuous on $\displaystyle [0,1]$ by Heine's theorem. As a consequence, $\displaystyle f$ is uniformly continuous on $\displaystyle (0,1]$ (this is weaker).