Differentiation to get $F'(0)$ I have $$f(x)=\begin{cases} 
      \sin\left(\frac{1}{x}\right) &\text{ if } x\neq 0, \\
      0 & \text{ if }x=0. 
   \end{cases}$$ and define $F(x)=\int_0^x f(t) dt$. I want to show that $F'(0)=f(0)$.
My idea is $$F'(0)=\lim_{h\to 0}\frac{F(h)-F(0)}{h}=\lim_{h\to 0}\frac{\int_0^h \sin\left(\frac{1}{x}\right) dx}{h}$$ and using a change of variables $t=\frac{1}{x}$ we have $$F'(0)=\lim_{h\to 0}\frac{\int_{\frac{1}{h}}^\infty\frac{\sin t}{t^2} dt}{h}.$$
I am not sure how to get this equal to $0$ now? Any help would be much appreciated.
 A: Consider
$$G(x) = \int_x^\infty \frac{\sin t}{t^2} \ dt$$ that is well defined on $[1, \infty)$. By integration by part, you get
$$G(x)=\frac{\cos x}{x^2} - 2 \int_x^\infty \frac{\cos t}{t^3} \ dt$$ and
$$\lim\limits_{x \to \infty} x G(x) = 0$$ as $\lim\limits_{x \to \infty}\frac{\cos x}{x} = 0$ and
$$\left\vert \int_x^\infty \frac{\cos t}{t^3} \ dt \right\vert \le  \int_x^\infty \frac{1}{t^3} \ dt = \frac{1}{2x^2}.$$
Then you indeed get the desired result
$$F^\prime(0)=f(0)=0$$ as
$$\lim_{h\to 0}\frac{\int_{\frac{1}{h}}^\infty\frac{\sin t}{t^2} dt}{h} = \lim\limits_{x \to \infty} x G(x) = 0.$$
A: Since
\begin{align}
\displaystyle\left|\dfrac{\int_0^h\sin\left(\frac{1}{x}\right)\mathrm{d}x}{h}\right|&=\left|\dfrac1h\int_0^h\sin\left(\frac{1}{x}\right)\mathrm{d}x\right|=\\
&=\left|\dfrac1h\int_0^h\!\left[D\left(x^2\cos\left(\dfrac1x\right)\right)-2x\cos\left(\frac{1}{x}\right)\right]\mathrm{d}x\right|\leqslant\\
&\leqslant\left|\dfrac1h\int_0^h\!\!D\left(x^2\cos\left(\dfrac1x\right)\right)\mathrm{d}x\right|+\left|\dfrac1h\int_0^h 2x\cos\left(\dfrac1x\right)\mathrm{d}x\right|=\\
&=\left|\dfrac1h\cdot h^2\cos\left(\dfrac1h\right)\right|+\left|\dfrac1h\int_0^h 2x\cos\left(\dfrac1x\right)\mathrm{d}x\right|\leqslant\\
&\leqslant\big|h\big|+\left|\dfrac1h\int_0^h\left|2x\cos\left(\dfrac1x\right)\right|\mathrm{d}x\right|\leqslant\\
&\leqslant\big|h\big|+\left|\dfrac1h\int_0^h \big|2x\big|\mathrm{d}x\right|=\\
&=\big|h\big|+\left|\dfrac1h\cdot\mathrm{sign}(h)\cdot h^2\right|=\\
&=2\big|h\big|\quad\text{ for all }\;h\in\mathbb{R}\;\land\;h\ne0\;,
\end{align}
there exists
$\lim\limits_{h\to0}\dfrac{\int_0^h\sin\left(\frac{1}{x}\right)\mathrm{d}x}{h}=0\;.$
