Finding $f'(0)$ when $f(x)=\int\limits_0^x\sin\left(\frac{1}{t}\right)dt$ I need to show that $f'(0)=0$ for
$$
f(x)=\int\limits_0^x\sin\left(\frac{1}{t}\right)dt
$$
But fundamental theorem of calculus is unapplicable here. What should I do?
 A: Here is one approach:


*

*Use integration by parts to show that $f(x)=x^2\cos(1/x)-\int_0^x 2t\cos(1/t)dt$ for $x\neq 0$.  

*Use this to show that $\left|\dfrac{f(x)}{x}\right|\leq 2|x|$.

A: To strengthen the convergence of an integral, a integration by parts is always a good idea.
Here we have
$$\begin{aligned}
f(x)
&=\int_0^x\sin\left(\frac 1t \right) dt\\
&= \left[ t^2 \cos\left(\frac 1t \right)\right]_0^x - \int_0^x 2t\cos\left(\frac 1t \right)dt\\
&= x^2 \cos\left(\frac 1x \right) - \int_0^x 2t\cos\left(\frac 1t \right)dt
\end{aligned}$$
You can apply the fundamental theorem of calculus for the second term, since it is the integral of a continuous function. The first term is a $O(x^2)$, so it is differentiable at zero with null derivative. In the end, we get $f'(0) = 0$.
[Reminder — A function $f$ is derivable at zero with derivative $a$ if and only if $f(x) = f(0) + ax + o(x)$ when $x\to 0$.]
A: You can use the fundamental theorem of calculus to show that $f$ is differentiable in an interval around $0$ and apply Rolle's theorem.
