Function is not continuous

Question

$f(x)=x^2 \sin(1/x)$ when $x \neq 0$, $f(x)=0$ when $x=0$.

Show that f is not continuously differentiable.

I found that $f'(x)=2x\sin(1/x)-\cos(1/x)$ for $x \neq 0$ and $0$ otherwise, but how can I prove this is not continuous?

Answer 1

If $f'(x)$ was continuous, its derivative at $0$ would be the value at $0$ of $Sin(1/x)$ (more precisely, the value at $0$ would agree with the limit of f'(x) as $x \rightarrow 0$). But , as you approach $0$, $Sin(1/x)$ oscillates between $-1$ and $1$. Use the fact that $Sin a$ has a period of $2\pi$ , so that $1/x$ will have period $1/2\pi$, then you can find sequences converging to $0$ that will take any value in $[-1,1]$. This violates the necessary condition for continuity that $Lim_{x\rightarrow x_0}f(x)=f(x_0)$.