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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

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$$

\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}

$$

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.

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