Question

If $f\left(x\right)=(x-[x\left]\right)\sin \frac{1}{x}$, then at $x=0$.

Answer 1

$\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}(x-0)\sin \frac{1}{x}=0$

$\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}(x+1)\sin \frac{1}{x}=\sin \left(\infty \right)$

So $\underset{x\to 0^{+}}{lim}f\left(x\right)\ne \underset{x\to 0^{-}}{lim}f\left(x\right)$

So at $x=0,f\left(x\right)$ is discontinuous