To find a limit involving integral How to find the following limit : $\lim _{x \to \infty} \dfrac 1 x \int_0^x \dfrac {dt}{1+x^2 \cos^2 t}$ ? I am not even sure whether the limit exists or not .  I tried applying L'Hospital , but then in the numerator we have differentiation under integration, and the derivative comes out to be lot messier than the original integral. Please help . Thanks in advance 
 A: Let 
$$F(x)=\frac {1}{x} \int_0^x \dfrac {dt}{1+x^2 \cos^2 t}$$
Then for $n\in\mathbb{Z}$, since $\cos^2(t)$ is $\pi$-periodic,
\begin{align*}
F(n\pi)
&=
\frac {1}{n\pi} \sum_{k=0}^{n-1}\int_{k\pi}^{(k+1)\pi} \dfrac {dt}{1+(n\pi)^2 \cos^2 t}\\
&=\frac {1}{n\pi} \sum_{k=0}^{n-1}\int_{0}^{\pi} \dfrac {dt}{1+(n\pi)^2 \cos^2 t}
=\frac {1}{\pi} \int_{0}^{\pi} \dfrac {dt}{1+(n\pi)^2 \cos^2 t}\\
&=\frac{1}{\sqrt{1+n^2\pi^2}}.
\end{align*}
which goes to zero as $n\to +\infty$.
Note that if $0<n\pi\leq x< (n+1)\pi$ then
$$0\leq F(x)\leq \frac {1}{n\pi} \int_0^{(n+1)\pi} \dfrac {dt}{1+(n\pi)^2 \cos^2 t}=F(n\pi)+\frac {1}{n\pi} \int_{0}^{\pi} \dfrac {dt}{1+(n\pi)^2 \cos^2 t}.$$ 
Can you take it from here?
A: $$\forall x > 0, \exists n \in \mathbb{N}\ s.t. \ x \in ]2n\pi, 2(n+1)\pi]$$
$$ 0 \leq \frac{1}{x}\int_0^x\frac{dt}{1+x^2\cos^2(t)}  \leq \frac{1}{2n\pi}\int_0^{2(n+1)\pi}\frac{dt}{1+(2n\pi)^2\cos^2(t)}$$
$$ \leq \frac{1}{2n\pi}\sum_{k=0}^n\int_0^{2(k+1)\pi}\frac{dt}{1+(2n\pi)^2\cos^2(t)}$$
$$ \leq \frac{1}{2n\pi}\sum_{k=0}^n\int_0^{2\pi}\frac{dt}{1+(2n\pi)^2\cos^2(t)}$$
$$ \leq \frac{1}{2n\pi}\sum_{k=0}^n4\int_0^{\pi/2}\frac{dt}{1+(2n\pi)^2\cos^2(t)}$$
$$ \leq \frac{2(n+1)}{n\pi}\int_0^{\pi/2}\frac{dt}{1+(2n\pi)^2\cos^2(t)}$$
$$ \leq \frac{2(n+1)}{n\pi}\int_0^{1}\frac{dt}{1+(2n\pi)^2t^2}$$
$$ \leq \frac{2(n+1)}{n\pi}\frac{\pi}{4n\pi} \to 0$$
