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$$\lim _{x\to \infty }\frac{1}{x}\int _0^x\:\frac{dt}{2+\cos t}$$

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 $\frac{1}{\sqrt{3}}$

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    Thats not an indeterminate to apply lhospital. Try half angle to convert it to tangent function. then wierstrass substitution then integrate it and get the result – 2017-02-20
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    You forgot one requirement of L'Hopital's rule: The limit of $f'(x)/g'(x)$ **must exist** in order to apply the rule and conclude that $\lim f/g = \lim f'/g'$. As you've noted, $\lim f'/g'$ doesn't exist in this case. This means that you can't apply L'Hopital's rule. It **doesn't** mean that $\lim f/g$ doesn't exist. – 2017-02-20
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    +1 for coming up with a limit which beats the L'Hospital's rule. The hypotheses of the rule are true here (denominator tends to $\infty$) but it is not helpful because the limit of $f'/g'$ does not exist. – 2017-02-21

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$f(t)=\frac{1}{2+\cos(t)}$ 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 $t=\arctan(u)$: $$ I = \frac{2}{\pi}\int_{0}^{+\infty}\frac{du}{3+u^2}=\color{red}{\frac{1}{\sqrt{3}}}.$$

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    where is the $1/x$ of the question? – 2017-02-20
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    @RafaelWagner: the (integral) mean value of a bounded, continuous, $2\pi$-periodic function is its mean value over $(0,2\pi)$, that is quite trivial. Typo fixed. – 2017-02-20
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    Where can I find proof that this works for a positive, bounded, $2\pi$. – 2017-02-20
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    @AhmedS.Attaalla Because it is bounded, and furthermore, periodic. Note that: $$\int_0^x\frac{dt}{2+\cos(t)}=ax'+\int_0^{x'}\frac{dt}{2+\cos(t)}$$where $a=\int_0^{2\pi}\frac{dt}{2+\cos(t)}$. and $x'=x\pmod{2\pi}$. – 2017-02-20
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    @AhmedS.Attaalla: the average value of $\cos x$ is zero hence the average value of $\cos(x)+2$ is $2$. Assume that $x=2\pi k+\tau$ and a $2\pi$-periodic bounded function has mean value over its period equal to $M$. Then $$\frac{1}{x}\int_{0}^{x}f(t)\,dt = \frac{1}{2\pi k+\tau}(2\pi k M+ B)$$ approaches $M$ as $x\to +\infty$. – 2017-02-20
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    Why is $\frac{1}{\pi}\int_{0}^{\pi}\frac{dt}{1+2\cos^2(t)}=\frac{2}{\pi}\int_{0}^{\pi/2}\frac{dt}{1+2\cos^2(t)}$ and how did you write the $\cos$ after substitution with $\arctan$ – 2017-02-20
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    @Liviu: $\cos^2(x)$ is a symmetric function with respect to $x=\frac{\pi}{2}$ and $\cos^2\arctan(t) = \frac{1}{t^2+1}$ is a useful trigonometric identity. – 2017-02-20
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    Thank you so much for responding to my question @JackD'Aurizio – 2017-02-20
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    You actually showed that a result is "quite trivial" by proving it in comments. It it were non-trivial it would need an answer. +1 for your answer as well as the comments dealing with the trivial result. – 2017-02-21
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    @JackD'Aurizio I stil don't understand the $\cos^2(x)$ part... I understood that $\cos(x)=\cos(-x)$ and that $\cos^2(x)$ is symmetric with respect to $x = \pi/2$ but why is that $2 \cdot$ in front of the integral ? Can you explain when this applies and how it's done, please ? – 2017-02-21
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    @Liviu: if $f\left(\frac{\pi}{2}-x\right)=f\left(\frac{\pi}{2}+x\right)$, then $$\int_{0}^{\pi} f(x)\,dx = 2 \int_{0}^{\pi/2} f(x)\,dx.$$ Draw a graph and shade some regions: it will become obvious. – 2017-02-21