$g: \mathbb{R} \to \mathbb{R} $
$$g(x) = \begin{cases} x^2\sin(1/x)& \text{if $x\ne 0$}, \\ 0 &\text{if $x = 0$}.\end{cases}$$
Prove that g is differentiable everywhere, and that its derivative $g':\mathbb{R} \to \mathbb{R}$ is continuous on $\{x\in\mathbb{R}| x\not=0\}$ but discontinuous at 0.
My attempt,
since $x^2$ is differentiable everywhere, and $\sin(1/x)$ is differentiable every $x\not=0$ then the product is differentiable everywhere but $x\not=0$. if $x = 0$ could I simply state that since $g(x) = 0$ then it is differentiable at $ x = 0$ or would I have to use $x^2\sin(1/x)$? I wrote down on paper that $-x^2 \leq x^2\sin(1/x) \leq x^2$ and used the sandwich theorem to say it's differentiable at x = 0. Is this true or would I have to use $g(x) = 0$?
I've done all other parts of the question, just stuck on this part.
1 Answer
By definition
$$g'(0)=\lim_{x\rightarrow 0}\frac{x^2\sin(\frac{1}{x})-g(0)}{x}=\lim_{x\rightarrow 0}\frac{x^2\sin(\frac{1}{x})}{x}= \lim_{x\rightarrow 0}x\sin(\frac{1}{x})=0 $$
by the sandwich theorem. We want to study continuity of
$$g'(x) = \begin{cases} 2x\sin(\frac{1}{x})-\cos(\frac{1}{x})& \text{if $x\ne 0$}, \\ 0 &\text{if $x = 0$}.\end{cases}.$$
Problems arise at $x=0$. Clearly the limits
$$\lim_{x\rightarrow 0^{\pm}}g'(x)$$
do not exist as $\cos(\frac{1}{x})$ admits no finite limit at $x=0$.
