Limit integral

As ∫(x→(x+pi)) |sinx| dx = 2 for all x,

∫(0→n) |sinx| dx = n*2/pi + constant error

The error is constant because the line y = 2x/pi is asymtopically equal to the integral f(x) = ∫(0→x) |sin u| du. f(x) - 2x/pi is periodic and checking the first period all points are definite.

Write

∫(0→n) |sinx| dx = 2n/pi + O(1) as x goes to infinity.

Now
lim(n→∞) ∫(1→n) |sinx|/n dx
= lim(n→∞)((1/n)( ∫(1→n) |sinx| dx ))
= lim(n→∞)((1/n)( ∫(0→n) |sinx| dx ) - (1/n)( ∫(0→1) |sinx| dx ))

As in the second term the integral converges but with 1/n in front of it it's certainly zero.

= lim(n→∞)((1/n)( ∫(0→n) |sinx| dx ))
= lim(n→∞)((1/n)(2n/pi + O(1)))
= lim(n→∞)(2/pi + O(n^-1))

As n -> infinity, O(n^-1) goes to zero.

= 2/pi.

If you don't like to use the big-O notation just sandwich the constant error and the result is the same.

Limit Inge 55=Brasil Use-----2015
Limit Inge 44=England Use 2014
Limit Inge 33=French Use 2013
Long Equal Inside Way Universe 2012