How to estimate the limit $\lim_{x\to+\infty}\frac{\int_0^x|\sin(s)|ds}{x}?$ I have come across the problem of estimation of the limit $$\lim_{x\to+\infty}\frac{\int_0^x|\sin(s)|ds}{x}.$$
Since it is easy to check that the method of l'Hopital's Rule is incapable of it. I have tried the following:  Since for every $x>\pi,$ there exist unique $n\in\mathbb{N}$ and $\theta\in[0, \pi),$ such that $$x=n\pi+\theta,$$
and $$x\to+\infty \Longleftrightarrow n\to\infty.$$
On account of the periodicity of the mapping $s\to |\sin(s)|,$ 
\begin{gather*}
\begin{aligned}
&\lim_{x\to+\infty}\frac{\int_0^x|\sin(s)|ds}{x}=\lim_{n\to\infty}\frac{\int_0^{n\pi+\theta}|\sin(s)|ds}{n\pi+\theta}=\lim_{n\to\infty}\frac{\int_0^{n\pi}|\sin(s)|ds+\int_{n\pi}^{n\pi+\theta}|\sin(s)|ds}{n\pi+\theta}\\
=&\lim_{n\to\infty}\frac{2n+\int_0^{\theta}|\sin(s)|ds}{n\pi+\theta}=\lim_{n\to\infty}\frac{2+\frac{1}{n}\cdot\int_0^{\theta}|\sin(s)|ds}{\pi+\frac{1}{n}\cdot\theta}\\
=&\frac{2}{\pi}.
\end{aligned}
\end{gather*}
I am not very sure that my trial is sound. Can anyone help me to check my method, or find another  way to estimate this limit? 
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$\ds{\lim_{x \to \infty}\bracks{{1 \over x}\int_{0}^{x}\verts{\sin\pars{s}}
     \,\dd s}:\ {\large ?}}$.
$$
\int_{0}^{x}\verts{\sin\pars{s}}\,\dd s
=
x\verts{\sin\pars{x}} - \int_{0}^{x}s\sgn\pars{\sin\pars{s}}\cos\pars{s}\,\dd s
=x\verts{\sin\pars{x}} - \int_{0}^{x}\sgn\pars{\sin\pars{s}}\varphi'\pars{s}\,\dd s
$$
where $\ds{\varphi\pars{s} = \int_{0}^{s}t\cos\pars{t}\,\dd t}$.
\begin{align}
\int_{0}^{x}\verts{\sin\pars{s}}\,\dd s&=
x\verts{\sin\pars{x}} - \sgn\pars{\sin\pars{x}}\varphi\pars{x}
+
\int_{0}^{x}\varphi\pars{s}\bracks{2\delta\pars{\sin\pars{s}}\cos\pars{s}}\,\dd s
\\[3mm]&=
x\verts{\sin\pars{x}} - \sgn\pars{\sin\pars{x}}\varphi\pars{x}
+
2\int_{0}^{x}\varphi\pars{s}\cos\pars{s}
\sum_{n = -\infty}^{\infty}\delta\pars{s - n\pi}\,\dd s
\\[3mm]&=
x\verts{\sin\pars{x}} - \sgn\pars{\sin\pars{x}}\varphi\pars{x}
+
2\sum_{n = 1}^{\infty}\pars{-1}^{n}\varphi\pars{n\pi}\Theta\pars{x - n\pi}
\end{align}
However,
$$
\varphi\pars{s} = s\sin\pars{s} - \int_{0}^{s}\sin\pars{t}\,\dd t
 = s\sin\pars{s} + \cos\pars{s} - 1\quad\imp\quad\varphi\pars{n\pi}
= \pars{-1}^{n} - 1
$$
\begin{align}\color{#0000ff}{\large%
\int_{0}^{x}\verts{\sin\pars{s}}\,\dd s
=
2\sin^{2}\pars{x \over 2}\sgn\pars{\sin\pars{x}}
+
4\sum_{n = 0}^{\infty}\Theta\pars{{x \over \pi} - 2n - 1}}
\end{align}

\begin{align}\color{#0000ff}{\large%
\lim_{x \to \infty}\bracks{{1 \over x}\int_{0}^{x}\verts{\sin\pars{s}}\,\dd s}}
&=
4\lim_{x \to \infty}\bracks{{1 \over x}\sum_{n = 0}^{\infty}
\Theta\pars{{x - \pi \over 2\pi} - n}}
\\[3mm]&=
4\lim_{x \to \infty}\bracks{{1 \over 2\pi x + \pi}\sum_{n = 0}^{\infty}
\Theta\pars{x - n}}
=
\color{#0000ff}{\large{2 \over \pi}}
\end{align}

