Determining whether $A(x) = \int_0^x |\sin(1/u) |\,du$ is differentiable at $0$ 
Let $$A(x) = \int_0^x |\sin(1/u) |\,du = \lim\limits_{a\to 0^+} \int_a^x |\sin(1/u)|\,du$$ Determine, with proof, whether $A$ is differentiable at $0$.

Clearly $A(0) = 0$. By definition, I need to show that $$\lim\limits_{x\to 0^+} \frac{A(x)}x = \lim\limits_{x\to 0^-} \dfrac{A(x)}x$$
I know $A$ is odd. Also,
$$A(x) = \sum_{k=k_x}^\infty \int_{1/((k+1)\pi)}^{1/(k\pi)} |\sin(1/u)| du + \int_{1/(k_x\pi)}^x |\sin(1/u)| du,$$
where $k_x = \lceil 1/(x\pi)\rceil$. Using the latter expression, I can show that
$$\left|\dfrac{A(x)}x\right|\leq \frac{1}{|x|}\left|\sum_{k=k_x}^\infty \frac{1}{k\pi} - \frac{1}{(k+1)\pi} + x-\frac{1}{k_x \pi}\right| = 1,$$ but obviously this isn't enough to show the derivative exists.
 A: The derivative  $A'(0)$ exists and is equal $2/\pi.$
By substitution $u=v^{-1}$ we get
$$A(x)=\int\limits_{1/x}^\infty {|\sin v|\over v^2}\,dv$$
Assume $${1\over (n+1)\pi}\le x\le {1\over n\pi}$$
Then
$$\int\limits_{(n+1)\pi}^\infty {|\sin v|\over v^2}\,dv\le A(x)\le \int\limits_{n\pi}^\infty {|\sin v|\over v^2}\,dv \quad (*)$$
We have
$$\int\limits_{k\pi}^\infty {|\sin v|\over v^2}\,dv = \sum_{l=k}^\infty \int\limits_{l\pi}^{(l+1)\pi}{|\sin v|\over v^2}\,dv\\ \le \sum_{l=k}^\infty {1\over l^2\pi^2} \int\limits_{l\pi}^{(l+1)\pi}|\sin v|\,dv =2 \sum_{l=k}^\infty {1\over l^2\pi^2} $$ Similarly we can show that
$$\int\limits_{k\pi}^\infty {|\sin v|\over v^2}\,dv = \sum_{l=k}^\infty \int\limits_{l\pi}^{(l+1)\pi}{|\sin v|\over v^2}\,dv\\ \ge \sum_{l=k}^\infty {1\over (l+1)^2\pi^2} \int\limits_{l\pi}^{(l+1)\pi}|\sin v|\,dv =2\sum_{l=k+1}^\infty {1\over l^2\pi^2} $$
Substituting  those estimates to $(*)$ implies
$$ {A(x)\over x}\le 2(n+1)\pi \sum_{l=n}^\infty {1\over  l^2\pi^2} ={2\over \pi}\, (n+1)\sum_{l=n}^\infty {1\over l^2},\\ \ \ \ 
{A(x)\over x}
\ge 2n\pi \sum_{l=n+1}^\infty {1\over  l^2\pi^2}={2\over \pi}\,n\sum_{l=n+1}^\infty {1\over l^2}$$
Next
$$\sum_{l=k}^\infty {1\over l^2}\le \sum_{l=k}^\infty {1\over (l-1)l}={1\over k-1},\\ \ \ 
\sum_{l=k}^\infty {1\over l^2}\ge \sum_{l=k}^\infty {1\over l(l+1)}={1\over k} $$
Therefore
$${2\over \pi} {n\over n+1}\le {A(x)\over x}\le {2\over \pi}{n+1\over n-1}$$
Hence
$$\lim_{x\to 0^+}{A(x)\over x}={2\over \pi}$$ which gives
$$A_+'(0)={2\over \pi}$$
