Prove $\lim_{x \rightarrow 0} \frac{f(x+g(x))-f(g(x))}{x}=f'(0)$ 
Given $f:\mathbb{R}\rightarrow \mathbb{R}$ differentiable continuously, and $g:\mathbb{R}\rightarrow \mathbb{R} \ s.t.\lim_{x \rightarrow 0}g(x)=0$ $$\text{Prove:} \lim_{x \rightarrow 0} \frac{f(x+g(x))-f(g(x))}{x}=f'(0) $$ 

$f'(x)$ is continuous $\Rightarrow\lim_{x \rightarrow 0}f'(g(x))=f'(\lim_{x \rightarrow 0}g(x))=f'(0) $  
By definition of the derivative: $\lim_{x \rightarrow 0} \frac{f(x+0)-f(0)}{x}=f'(0)$
How can I continue to get the result?
Any help appreciated.
 A: Since $f'$ is continuous, we may write first :
$$f(g(x)+x)-f(g(x))=\int_{g(x)}^{g(x)+x}f'(t)\,dt$$
then, for all $x\neq 0$ :
$$\frac{f(g(x)+x)-f(g(x))}{x}-f'(0)=\frac{1}{x}\int_{g(x)}^{g(x)+x}\left(f'(t)-f'(0)\right)\,dt$$
Now, given $\epsilon>0$, there exists $\eta>0$ such that :
$$|t|\le\eta\implies|f'(t)-f'(0)|\le\epsilon$$
Since $\lim_{x\to0}g(x)=\lim_{x\to0}\left(g(x)+x\right)=0$, there exists $\alpha>0$ such that :
$$|x|\le\alpha\implies\left(|g(x)|\le\eta\quad\mathrm{and}\quad|g(x)+x|\le\eta\right)$$
Hence, as soon $|x|\le\alpha$ :
$$\left|\frac{f(g(x)+x)-f(g(x))}{x}-f'(0)\right|\le\frac{1}{|x|}|x|\epsilon=\epsilon$$

Another proof
For all nonzero $x$, there exists $c_x\in\left(g(x),g(x)+x\right)$ or $\left(g(x)+x,g(x)\right)$ (depending on whether $x>0$ or $x<0$) such that :
$$f\left(g(x)+x\right)-f\left(g(x)\right)=\left((g(x)+x)-g(x)\right)\,f'(c_x)=x\,f'(c_x)$$
By squeeze theorem, we have $\lim_{x\to0}c_x=0$ because $\lim_{x\to0}g(x)=0$.
Hence, since $f'$ is continuous : $\lim_{x\to0}f'(c_x)=f'(0)$; in other words :
$$\lim_{x\to0}\frac{f\left(g(x)+x\right)-f\left(g(x)\right)}{x}=f'(0)$$
