$f(x)=x^2\sin(1/x+1),x >0$ $f(x)= 0, x \le 0$ continuously differentiable, differentiable twice?

Question

$f(x)=x^2 \sin(1/x+1),x >0$

$f(x)= 0, x \le 0$

Is $f(x)$ continuously differentiable? And is $f(x)$ differentiable twice?

How can I check/proof this?

My try was determining

$f'(x)=2x \sin(1/x)-\cos(1+1/x), x>0$ and

$f'(x)=0, x \le 0$

but what to do next?

Answer 1

it is $$\frac{f(0+h)-f(0)}{h}=\frac{h^2\sin(1/h+1)}{h}$$ and you must compute $$\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}$$ if this Limit exists and we get $$\left|h\sin(1+\frac{1}{h})\right|\le |h|$$ and this tends to Zero if $h$ tends to zero