A question about differentiation $g(x)=\sin(1/x)  ,  g(0)=0$
Define $G(x)=\int _{0}^{x}g\left( t\right) dt$
Show $G'(0)=g(0)$
I use the definition
$$G'(0) = \lim _{h\rightarrow 0}\dfrac {G\left( h\right) -G\left( 0\right) } {h}$$
$$=\lim _{h\rightarrow 0}\dfrac {\int _{0}^{h}g\left( t\right) dt} {h}$$
$$=\lim _{h\rightarrow 0}\dfrac {\int _{0}^{h}\sin(1/t) dt} {h}$$
I have no idea about next step 
Can you give me some hint
Thank you!!!
 A: Expanding my comment, let $F(x) =x^{2}\cos(1/x),F(0)=0,f(x)=2x\cos(1/x),f(0)=0$. We can observe that $F$ is differentiable everywhere, $f$ is continuous everywhere and $g$ is continuous everywhere except at $x=0$ and $g$ is bounded. It is easily seen that $F'(x) =f(x)+g(x) $ for all $x$ and $f(x) +g(x) $ is Riemann integrable on any closed and bounded interval. Thus by second fundamental theorem of calculus we have $$F(x) - F(0)=\int_{0}^{x}\{f(t)+g(t)\}\,dt$$ or $$x^{2}\cos\left(\frac{1}{x}\right)=\int_{0}^{x}f(t)\,dt+\int_{0}^{x}g(t)\,dt$$ or $$x\cos\left(\frac{1}{x}\right)= \frac{1}{x}\int_{0}^{x}f(t)\,dt+\frac{1}{x}\int_{0}^{x}g(t)\,dt$$ Taking limits as as $x\to 0$ we get $$0=f(0)+\lim_{x\to 0}\frac{1}{x}\int_{0}^{x}g(t)\,dt$$ where the limit for the first term on right evaluates to $f(0)$ because of the first fundamental theorem of calculus and the fact that $f$ is continuous at $x=0$. Since $f(0)=0$, it follows that $$G'(0)=\lim_{x\to 0}\frac{1}{x}\int_{0}^{x}g(t)\,dt=0=g(0)$$
A: Use L'Hopital! 

Remembering that $${d\over dx}\int _{f(x)}^{g(x)}h(t)dt=g'(x)h(g(x))-f'(x)h(f(x))$$

