Looking for an elementary solution of this limit I was collecting some exercises for my students, and I found this one in a book: compute, if it exists, the limit
$$
\lim_{x \to +\infty} \int_x^{2x} \sin \left( \frac{1}{t} \right) \, dt.
$$
It seems to me that this limit exists by monotonicity. Moreover, since $\frac{2}{\pi}x \leq \sin x \leq x$ for $0 \leq x \leq \pi/2$, I could easily show that
$$
\frac{2}{\pi} \log 2 \leq \lim_{x \to +\infty} \int_x^{2x} \sin \left( \frac{1}{t} \right) \, dt \leq \log 2.
$$
WolframAlpha suggests a "closed" form for the integral, and by dominated convergence the limit turns out to be $\log 2$. However, passing to a limit in the $\operatorname{Ci}(\cdot)$ function is not really elementary. I wonder if there is a simpler approach that a student can understand at the end of a first course in mathematical analysis.

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\begin{align}
\lim_{x \to \infty}\int_{x}^{2x}\sin\pars{1 \over t}\,\dd t&=
\lim_{x \to \infty}\int_{1/x}^{1/2x}\sin\pars{t}\,\pars{-\,{\dd t \over t^{2}}}
=\lim_{x \to \infty}\int_{1/x}^{1/2x}{\cos\pars{t} \over t}\,\dd t
\\[3mm]&=\lim_{x \to \infty}\braces{%
-\bracks{-\int_{1/x}^{\infty}{\cos\pars{t} \over t}\,\dd t}
+
\bracks{-\int_{1/2x}^{\infty}{\cos\pars{t} \over t}\,\dd t}}
\\[3mm]&=
\lim_{x \to \infty}\bracks{-{\rm ci}\pars{1 \over x} + {\rm ci}\pars{1 \over 2x}}
\end{align}
where ${\rm ci}\pars{z}$ is the $\it\mbox{Cosine Integral}$ which satisfies
$\lim_{z \to 0}\bracks{{\rm ci}\pars{z} - \gamma - \ln\pars{z}} = 0$. $\gamma$ is the Euler-Mascheroni constant. Then,

\begin{align}
&\lim_{x \to \infty}\int_{x}^{2x}\sin\pars{1 \over t}\,\dd t=
\lim_{x \to \infty}\braces{%
-\bracks{{\rm ci}\pars{1 \over x} - \gamma - \ln\pars{1 \over x}}
+\bracks{{\rm ci}\pars{1 \over 2x} - \gamma - \ln\pars{1 \over 2x}} + \ln\pars{2}}
\end{align}

$$\color{#00f}{\large%
\lim_{x \to \infty}\int_{x}^{2x}\sin\pars{1 \over t}\,\dd t = \ln\pars{2}}
$$
