$$f(x) =
\begin{cases}
x\sin(1/x), & \text{if $x$ $\ne$ $0$} \\
0, & \text{if $x$ = $0$} \\
\end{cases}$$
Is $f$ continuous on $(-1/\pi$, 1/$\pi$)?
Is $f$ differentiable on $(-1/\pi$, 1/$\pi$)?
I know how to prove continuity on a single point, but I'm not sure how to prove continuity for a whole interval. Also, I know there is a theorem that states that if a function is differentiable at a point, then it's continuous.
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