Let $f(x) = x^2 \sin{\frac{1}{x}}$ for $x\neq 0$ and $f(0) =0$.

(a) Use the basic properties of the derivative, and the Chain Rule to show that $f$ is differentiable at each $a\neq 0$ and calculate $f'(a)$.

You may use without proof that $\sin$ is differentiable and that $\sin' =\cos$.

Not even sure what this is asking.

(b) Show that $f$ is differentiable at $0$ and that $f'(0) =0$.

$\frac {f(x)-f(0)}{x-0} \to \lim_{x \to 0} x \sin(1/x)$.

$x \sin(1/x) \leq |x|$ and $\lim_{x \to 0} |x|=0$.

Thus $f(x)$ is differentiable at $0$; moreover $f^{'}(0)=0$.

(c) Show that $f'$ is not continuous at $0$.

$f{'}(x)=x^{2} \cos(1/x) (-x^{-2}) + 2x \sin (1/x)$.

In pieces: $\lim_{x \to 0} \cos (1/x)$.

$f^{'}(0-)$ nor $f{'}(0+)$ exists as $x \to 0$ $f^{'}(x)$ oscillates infinity between $-1$ and $1$ with ever increase frequency as $x \rightarrow 0$ for any $p>0$ $[-p,0]$, $[-p,p]$ or $[0,p]$ $f$ is not continuous.

**Question**: How to show more rigorously?