$f'(0)$ when $f(x)=x^2\sin(1/x)$

Question

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

This is presented as an MCQ and answer is given as "$f$ is differentiable at every $ x$ but $f'$ is discontinous at $0$ "

I calculated LHD and RHD at $0$ as $0$ and $f'(x)=2x\sin(1/x)-\cos(1/x)$ which is not defined at 0 thus the answer should be "$f$ is not differentiable at $0$"

Is this a misprint or am I missing something?

Answer 1

Yes, it is defined at $0$, since$$f'(0)=\lim_{x\to0}\frac{x^2\sin\left(\frac1x\right)}x=\lim_{x\to0}x\sin\left(\frac1x\right)=0.$$What isn't defined at $0$ is $\lim_{x\to0}f'(x)$, but that's another matter.

Answer 2

Use that $$f'(x_0)=\lim_{h\to 0}\frac{f(x_0+h)-f(x_0)}{h}$$ and let $$x_0=0$$ then we have $$\frac{f(0+h)-f(0)}{h}=\frac{h^2\sin(\frac{1}{h})}{h}=h\sin(\frac{1}{h})$$ so $$\lim_{h\to 0}h\sin(\frac{1}{h})=0$$