TITLE: Evaluate limit of an integral: $\lim_{x\to \infty }\frac{1}{x}\int _0^x\:\frac{dt}{2+\cos t}$
QUESTION [9 upvotes]: $$\lim _{x\to \infty }\frac{1}{x}\int _0^x\:\frac{dt}{2+\cos t}$$
Can someone explain to me if it is a limit of type $\frac{\infty}{\infty}$ or not and why ? I considered it to be one, applied L'Hospital and got $\cos\infty$, which would mean that the limit does not exist, but the answer is $\frac{1}{\sqrt{3}}$

REPLY [9 votes]: $f(t)=\frac{1}{2+\cos(t)}$ is a positive, bounded and $2\pi$-periodic function. It follows that the mean value of $f$, i.e. the wanted limit, equals the mean value of $f$ over one period:
$$ \lim_{x\to +\infty}\frac{1}{x}\int_{0}^{x}\frac{dt}{2+\cos t} = \frac{1}{2\pi}\int_{0}^{2\pi}\frac{dt}{2+\cos t} = I $$
and through the substitution $t\mapsto 2t$ we have:
$$ I = \frac{1}{\pi}\int_{0}^{\pi}\frac{dt}{1+2\cos^2(t)}=\frac{2}{\pi}\int_{0}^{\pi/2}\frac{dt}{1+2\cos^2(t)} $$
then by setting $t=\arctan(u)$:
$$ I = \frac{2}{\pi}\int_{0}^{+\infty}\frac{du}{3+u^2}=\color{red}{\frac{1}{\sqrt{3}}}.$$