# 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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Very nicely done, and excellently presented. –  Stephen Montgomery-Smith Jan 3 '14 at 1:00
Thanks, Stephen Montgomery-Smith! Your comment gives me confidence. –  nuage Jan 3 '14 at 1:11
Having a dangling free $\theta$ in your limit doesn't make sense. One way to fix it is define $n$ and $\theta$ as a short hand of $\lfloor\frac{x}{\pi}\rfloor$ and $x - n\pi$ and then take the limit in $x$ instead of $n$. –  achille hui Jan 3 '14 at 1:30
Since $\theta$ is always in $[0,\pi)$, and $\frac{1}{n}\to 0,$ as $n\to\infty,$ we have $\frac{1}{n}\cdot \theta\to 0,$ as $n\to\infty.$ –  nuage Jan 3 '14 at 1:55
It seems like it's only a notational problem and achille hui commented on how to fix that. The idea and the answer is still right. –  Pratyush Sarkar Jan 3 '14 at 2:47
