Proof that $g(x) = xf(x)$ is differentiable at 0 given that f is continuous at 0

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Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be continous at $0$ and let $g : \mathbb{R} \rightarrow \mathbb{R}$ be $g(x)=xf(x)$ for all $x\in \mathbb{R}$. How do I proof $g$ is differentiable at $0$?

I know that the product of two differentiable functions is differentiable, and that $x$ is differentiable. So I guess I need to proof that $f(x)$ is differentiable at $0$? Continuity does however not imply differentiability, so I wouldn't know how.

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Consider $$\frac{g(h)-g(0)}{h}=\frac{hf(h)}h=f(h).$$ For $g'(0)$ to exist, all one need is for this to tend to a limit as $h\to0$.