Derivative in 0 I'm a highschool student and we don't learn maths in English. So please excuse me for my Math's English.
I'm doing an exercise and I can't answer its final question. Can you help me? Thank you!
Let $f$ a differentiable function on $\mathbb{R}$ which derivative is continuous on $\mathbb{R}$.
Let $g$ a function defined on $\mathbb{R}$ by:
$\forall x\in\mathbb{R}^*,\,\,g(x)=\dfrac{1}{2x}\int_{-x}^xf(t)\,dt$ and $g(0)=f(0)$.
1- Prove that $g$ is continuous at $0$.
2- For all $x\ne 0$ calculate $g'(x)$ according to $f(x)$ and $g(x)$.
I answerd both questions. For the seconde one I found:
$$\forall x\in\mathbb{R}^*,\,\,g'(x)=\dfrac{f(x)+f(-x)-2g(x)}{2x}$$
Here's the last question.
3- Prove that $g$ is differentiable at $0$ and that $g'(0)=0$.
Thank you.
edit: That's what I tried:
$\lim_{x\to 0}\dfrac{g(x)-g(0)}{x}=\lim_{x\to 0}\dfrac{1}{2}\left(\dfrac{\frac{F(x)}{x}-f(0)}{x}+\dfrac{\frac{F(-x)}{-x}-f(0)}{x}\right) $
where $F(x)=\int_0^xf(t)dt$

We have that
$$g'(0)=\lim_{x\to 0}\frac{\int_{-x}^x f(t)dt-2f(0)x}{2x^2}=\lim_{x\to 0}\frac{f(x)+f(-x)-2f(0)}{4x}$$ $$=\lim_{x\to 0}\frac{f'(x)-f'(-x)}{4}=\frac{f'(0)-f'(0)}{4}=0,$$
where we have applied L'Hospital rule in the second and third equalities.
A clarification
Note that because of the fundamental theorem of calculus $F(x)=\int_0^xf(t)dt$ is differentiable at any point where $f$ is continuous and $F'(x)=f(x).$ So $\int_{-x}^xf(t)dt=F(x)-F(-x)$ is differentiable at any point and the same happens to $\int_{-x}^xf(t)dt-2f(0)x.$ For this reason we can apply L'Hospital rule in the second equality. 
