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 

$$\forall x > 0, \exists n \in \mathbb{N}\ s.t. \ x \in ]2n\pi, 2(n+1)\pi]$$
$$ 0 \leq \frac{1}{x}\int_0^x\frac{dt}{1+x^2\cos^2(t)}  \leq \frac{1}{2n\pi}\int_0^{2(n+1)\pi}\frac{dt}{1+(2n\pi)^2\cos^2(t)}$$
$$ \leq \frac{1}{2n\pi}\sum_{k=0}^n\int_0^{2(k+1)\pi}\frac{dt}{1+(2n\pi)^2\cos^2(t)}$$
$$ \leq \frac{1}{2n\pi}\sum_{k=0}^n\int_0^{2\pi}\frac{dt}{1+(2n\pi)^2\cos^2(t)}$$
$$ \leq \frac{1}{2n\pi}\sum_{k=0}^n4\int_0^{\pi/2}\frac{dt}{1+(2n\pi)^2\cos^2(t)}$$
$$ \leq \frac{2(n+1)}{n\pi}\int_0^{\pi/2}\frac{dt}{1+(2n\pi)^2\cos^2(t)}$$
$$ \leq \frac{2(n+1)}{n\pi}\int_0^{1}\frac{dt}{1+(2n\pi)^2t^2}$$
$$ \leq \frac{2(n+1)}{n\pi}\frac{\pi}{4n\pi} \to 0$$
