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.

$$
\int_x^{2x}\sin\left(\frac{1}{t} \right)\,dt=\int_{\frac{1}{2x}}^{\frac{1}{x}}\frac{\sin(u)}{u^2}\,du,
$$
$$
\sin(u)=u+o(1)u,
$$
if $u$ is around $0$,
(here we use the elemetary $\lim_{u\to0}\frac{\sin(u)}{u}=1$) 
so
$$
\int_{\frac{1}{2x}}^{\frac{1}{x}}\frac{\sin(u)}{u^2}\,du=\int_{\frac{1}{2x}}^{\frac{1}{x}}\frac{1}{u}(1+o(1))\,du,
$$
where $o(1)\to 0$ as $u\to0$, which is the case, because $x\to+\infty$.
