Does the limit $\lim _{x\to 0} \frac 1x \int_0^x \left|\cos \frac 1t \right| dt$ exists?

Solution 1:

$$ \begin{align} \lim_{x\to0}\frac1x\int_0^x\left|\,\cos\left(\frac1t\right)\,\right|\mathrm{d}t &=\lim_{x\to\infty}x\int_x^\infty\frac{|\cos(t)|}{t^2}\,\mathrm{d}t\\ &=\lim_{n\to\infty}(x+n\pi)\sum_{k=n}^\infty\int_x^{x+\pi}\frac{|\cos(t)|}{(t+k\pi)^2}\,\mathrm{d}t\tag{1} \end{align} $$ Using the Euler-Maclaurin Sum Formula, $$ \sum_{k=n}^\infty\frac{x+n\pi}{(t+k\pi)^2} =\frac{x+n\pi}{\pi(t+n\pi)}+\frac{x+n\pi}{2(t+n\pi)^2}+O\!\left(\frac1{n^2}\right)\tag{2} $$ Thus, for $x\le t\le x+\pi$,we have $$ \sum_{k=n}^\infty\frac{x+n\pi}{(t+k\pi)^2}=\frac1\pi+O\!\left(\frac1n\right)\tag{3} $$ and therefore, $$ \begin{align} \lim_{x\to0}\frac1x\int_0^x\left|\,\cos\left(\frac1t\right)\,\right|\mathrm{d}t &=\frac1\pi\int_0^\pi|\cos(t)|\,\mathrm{d}t\\ &=\frac2\pi\tag{4} \end{align} $$

Solution 2:

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

1. 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 $$ \frac{2}{(x+\pi(k+1))^2}\le\int_{x+\pi k}^{x+\pi(k+1)}\frac{|\cos t|}{t^2}\,dt\le\frac{2}{(x+\pi k)^2}. $$
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 $$ \frac{2}{\pi}\int_{x+\pi}^{\infty}\frac{1}{t^2}\,dt\le \int_x^{\infty}\frac{|\cos t|}{t^2}\,dt\le \frac{2}{\pi}\int_{x-\pi}^{\infty}\frac{1}{t^2}\,dt. $$
3. 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.