Does the limit $\lim _{x\to 0} \frac 1x \int_0^x \left|\cos \frac 1t \right| dt$ exists? Does the limit $\lim _{x\to 0} \frac 1x \int_0^x \left|\cos \frac 1t \right| dt$ exists ? If it does then what is the value ? 
I don't think even L'Hospital's rule can be applied . Please help . Thanks in advance 

As en alternative for the powerfull Euler-Maclaurin asymptotics in @robjohn answer one can use simple zero order approximations and the Squeeze theorem.


*

*The estimate for $x+\pi k\le t\le x+\pi(k+1)$ 
$$
\frac{|\cos t|}{(x+\pi(k+1))^2}\le \frac{|\cos t|}{t^2}\le \frac{|\cos t|}{(x+\pi k)^2}
$$
and integration gives 
$$
\frac{2}{(x+\pi(k+1))^2}\le\int_{x+\pi k}^{x+\pi(k+1)}\frac{|\cos t|}{t^2}\,dt\le\frac{2}{(x+\pi k)^2}.
$$

*Further estimation by the integrals
$$
\frac{2}{\pi}\int_{x+\pi(k+1)}^{x+\pi(k+2)}\frac{1}{t^2}\,dt\le \frac{2}{(x+\pi(k+1))^2}\le\ldots\le 
\frac{2}{(x+\pi k)^2}\le \frac{2}{\pi}\int_{x+\pi(k-1)}^{x+\pi k}\frac{1}{t^2}\,dt
$$
and summing up for $0\le k<+\infty$ gives
$$
\frac{2}{\pi}\int_{x+\pi}^{\infty}\frac{1}{t^2}\,dt\le \int_x^{\infty}\frac{|\cos t|}{t^2}\,dt\le \frac{2}{\pi}\int_{x-\pi}^{\infty}\frac{1}{t^2}\,dt.
$$

*Calculate the integral estimates and multiply by $x$
$$
\frac{2}{\pi}\frac{x}{x+\pi}\le x\int_x^{\infty}\frac{|\cos t|}{t^2}\,dt\le \frac{2}{\pi}\frac{x}{x-\pi}.
$$
Take the limit as $x\to\infty$ and use the Squeeze theorem.

