Evaluating $\lim_{x \to \infty}\frac{1}{x}\int_0^x|\sin(t)|dt$ I would appreciate some help with this problem:
Evaluate:
$$\lim_{x \to \infty}\frac{1}{x}\displaystyle\int_0^x|\sin(t)|dt$$

Ok, thanks for your comments I think I get it now:
$\displaystyle\int_0^{2\pi}|\sin (t)|dt$ = 4
And due to the periodicity of $|\sin (t|)|$, 
$\displaystyle\int_0^{2k\pi}|\sin (t)|dt = k\displaystyle\int_0^{2\pi}|\sin (t)|dt = 4k$
So the limit is equivalent to:
$$\lim_{k\to\infty}\left(\frac{1}{2k\pi}\cdot k\displaystyle\int_0^{2\pi}|\sin (t)|dt\right) = \lim_{k\to\infty}\frac{4k}{2k\pi} = \frac{2}{\pi}$$
... so in reality this question probably wasn't as hard as I first thought it was :/
