Behaviour of $x^2\sin(\frac{1}{x})$ at $x=0$

Question

Problem : Consider the function $f(x) = x^2\sin(\frac{1}{x})$. A priori this function is defined for $x\neq 0$. Show that setting $f(0) = 0$ turns $f$ into a function that is differentiable everywhere.

I've no idea where to start or how to go about tackling this problem at all.

My current idea is to prove that as $x$ tends to $0$, the limit of the function $= 0$, but after this I do not know how to continue.

I've also tried looking at it with the Taylor's Theorem but I do not know how to start off with this as well.

Thanks for your help.

Answer 1

Hint: For $x \ne 0$, you can simply use the standard rules of computing derivatives. For $x = 0$, use the definition of the derivative i.e. $$ \lim_{a \to 0} \frac{f(a) - f(0)}{a - 0}. $$

Answer 2

Compute $$\lim_{h\to 0}\frac{f(x_0+h)-f(x_0)}{h}$$ and let $$x_0=0$$ and we get $$\frac{f(h)-f(0)}{h}=\frac{h^2\sin(\frac{1}{h})}{h}=h\sin(\frac{1}{h})$$ and $$|h\sin(\frac{1}{h})|\le |h|$$ and this tends to $0$ for $h$ tends to $0$