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.
 A: Integrate by parts, with $u=\sin\left(\frac{1}{t}\right)$ and $dv=dt$.  We get that this is equal to 
$$
\left[t\sin\left(\frac{1}{t}\right)\right|_x^{2x}+\int_x^{2x}\frac{1}{t}\cos\left(\frac{1}{t}\right)dt
$$
Now the first term equals 
$$
\frac{\sin\left(\frac{1}{2x}\right)}{\frac{1}{2x}}-\frac{\sin\left(\frac{1}{x}\right)}{\frac{1}{x}}
$$
which limits to zero.  On the second term, for sufficiently large $t$, $1-\varepsilon\leq\cos\left(\frac{1}{t}\right)\leq 1$.  So we get that the second term is between $(1-\varepsilon)\log2$ and $\log2$.
A: $$\int_{x}^{2x}{\sin\left(\frac{1}{t}\right)dt}=\int_{x}^{2x}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)!}\cdot t^{-(2n+1)}\,dt= \sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)!} \int_{x}^{2x}{t^{-(2n+1)} dt}$$
Since 
$$\lim_{x \rightarrow \infty} \int_{x}^{2x}{t^{-(2n+1)}dt}=\begin{cases}0,& n >0
\\ \log2,& n=0\end{cases} $$
We conclude that $$\lim_{x \rightarrow \infty} \int_{x}^{2x}{\sin\left(\frac{1}{t}\right)\,dt}= \log2.$$
A: $$
\int_x^{2x}\sin\left(\frac{1}{t} \right)\,dt=\int_{\frac{1}{2x}}^{\frac{1}{x}}\frac{\sin(u)}{u^2}\,du,
$$
$$
\sin(u)=u+o(1)u,
$$
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 $x\to+\infty$.
A: $\newcommand{\+}{^{\dagger}}%
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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}

$$\color{#00f}{\large%
\lim_{x \to \infty}\int_{x}^{2x}\sin\pars{1 \over t}\,\dd t = \ln\pars{2}}
$$
A: Hint:
$$\int_{x}^{2x}\sin\left(\frac{1}{t}\right)\,dt=\frac{1}{2}\int_{2x}^{4x}\sin\left(\frac{2}{t}\right)\,dt=\int_{2x}^{4x}\sin\left(\frac{1}{t}\right)\cos\left(\frac{1}{t}\right)\,dt<\int_{2x}^{4x}\sin\left(\frac{1}{t}\right)\,dt$$
A: In the continuation of what Gaffney wrote, Cos[1/t] can be developed for large values of t as     
1 - 1/(2 t^2) + 1/(24 t^4) - 1/(720 t^6) + ....  
So, the value of Gaffney's last integral between $x$ and $2x$ is    
Log[2] - 3/(16 x^2) + 5/(512 x^4) + ...
A: If $t$ approaches infinity, $sin(1/t)$ will approach $0$, so you can just use the Taylor expansion, which we know to be absolutely convergent (if the students have had Taylor series in their first analysis course):
$$ \begin{align*}& lim_{x\to\infty} \int_x^{2x} sin(1/t) dx\\
= & lim_{x\to\infty} \int_x^{2x} (1/t) + O((1/t)^3) dx\\
= & lim_{x\to\infty} \left[ln(t) + O((1/t)^2)\right]_x^{2x}\\
= & lim_{x\to\infty} \left[ln(2) + O((1/x)^2)\right]\\
= & log(2)
\end{align*} $$
