If $f(0) = 0$, then, is $\mathrm{lim}_{x\to0} xf'(x) = 0$?

Is the above claim true? I think that it is. I am assuming that $f$ is continuous and differentiable at $0$. My not-so-rigorous attempt:

$$ \mathrm{lim}_{x \to 0} x f'(x) = \mathrm{lim}_{x \to 0} \mathrm{lim}_{h\to0} x \frac{f(h)}{h} $$

Now choosing the approach direction $x=h$, we see that the limit equals $0$ because $f(0)=0$. Is the idea correct? If not, can this condition be true under some additional constraints?

Assuming that $f$ is differentiable is not sufficient.

For example, set $f(x) = x^2 \sin(x^{-2})$ if $x \ne 0$, and $f(0)=0$.

Then $f'(0) = \lim_{h \to 0} \frac{f(h)-f(0)}{h} = \lim_{h \to 0} h \sin(h^{-2}) = 0$.

But when $x \ne 0$, $f'(x) = 2x \sin(x^{-2}) - x^2 \times 2x^{-3} \sin(x^{-2})$, so $xf'(x) = 2x^2 \sin(x^{-2}) - 2 \sin(x^{-2})$, which has no limit as $x \to 0$.