Computing the limit of an integral with the Squeeze theorem I am trying to compute the limit of the sequence $a_n$, as n approaches infinity:
$$a_n=\int_n^{2n} \frac{1}{x}e^\frac{1}{x} \,dx$$
Here's what I managed to do:
It is known that $ x+1\le e^x $ for any real $x$. Therefore, $\frac{1}{x}+1<e^\frac{1}{x}$. This means that $\frac{1}{x^2}+\frac{1}{x}<\frac{1}{x}e^\frac{1}{x}$. Integrating from n to 2n, we get that:
$$\int_n^{2n} (\frac{1}{x^2}+\frac{1}{x}) \,dx\leqslant a_{n}$$
Computing the integral, we obtain $\frac{1}{2n}+\ln(2)\leqslant a_n$
I decided to compute the integral numerically for some small values of n, using computer software. It looked like the limit was $\ln(2)$.
With that in mind, I tried to find an upper bound for $a_n$ so I can use the Squeeze theorem, but it was to no avail. Any ideas? Thanks!
 A: Let $ n $ be a positive integer.
Substitute : $\small \left\lbrace\begin{aligned}u&=\frac{1}{x}\\ \mathrm{d}x&=-\frac{\mathrm{d}u}{u^{2}}\end{aligned}\right. $, we get : \begin{aligned} \int_{n}^{2n}{\frac{\mathrm{e}^{\frac{1}{x}}}{x}\,\mathrm{d}x}=\int_{\frac{1}{2n}}^{\frac{1}{n}}{\frac{\mathrm{e}^{x}}{x}\,\mathrm{d}x}&=\int_{\frac{1}{2n}}^{\frac{1}{n}}{\frac{\mathrm{e}^{x}-1}{x}\,\mathrm{d}x}+\ln{2}\\ &=\int_{0}^{\frac{1}{n}}{\frac{\mathrm{e}^{x}-1}{x}\,\mathrm{d}x}-\int_{0}^{\frac{1}{2n}}{\frac{\mathrm{e}^{x}-1}{x}\,\mathrm{d}x}+\ln{2} \\ &=\int_{0}^{\frac{1}{n}}{\frac{\mathrm{e}^{x}-1}{x}\,\mathrm{d}x}-\int_{0}^{\frac{1}{n}}{\frac{\mathrm{e}^{\frac{x}{2}}-1}{x}\,\mathrm{d}x}+\ln{2}\\ a_{n}&=\int_{0}^{\frac{1}{n}}{\frac{\mathrm{e}^{x}-\mathrm{e}^{\frac{x}{2}}}{x}\,\mathrm{d}x}+\ln{2}\end{aligned}
In the second line we broke the integral apart, in deed we are allowed to, because $ f:x\mapsto\frac{\mathrm{e}^{x}-1}{x} $ is continuous in $ \left(0,+\infty\right) $, and can be extended into a countinuous function on $ \left[0,+\infty\right[ $, thus $ f $ is piecewise continuous on $ \left[0,+\infty\right[ $, hence it is integrable on $ \left[0,+\infty\right[ $, and on every segment of $ \left[0,+\infty\right[ \cdot $
In the third line we used a substitution $ \left(u=2x\right) \cdot $
Now since $ g:x\mapsto\frac{\mathrm{e}^{x}-\mathrm{e}^{\frac{x}{2}}}{x} $ can also be extended into some continuous function on $ \left[0,+\infty\right[ $, that means that $ g $ is bounded, on $ \left]0,1\right] $ in particular : There exists some $ M\in\mathbb{R}_{+} $, such that : $ \left(\forall x\in\left[0,1\right]\right),\ \left|g\left(x\right)\right|\leq M \cdot $
Thus : $$ \left(\forall n\in\mathbb{N}^{*}\right),\ \left|\int_{0}^{\frac{1}{n}}{\frac{\mathrm{e}^{x}-\mathrm{e}^{\frac{x}{2}}}{x}\,\mathrm{d}x}\right|\leq\frac{M}{n}\underset{n\to +\infty}{\longrightarrow}0 $$
Hence : $$ \lim_{n\to +\infty}{a_{n}}=\ln{2} $$
A: Option: MVT of integration.
$a_n=e^{1/t}\displaystyle {\int_{n}^{2n}}(1/x)dx$,
where $t \in [n,2n]$.
$a_n=e^{1/t}(\log 2n-\log n)=$
$e^{1/t}\log 2$.
$e^{1/2n}\log 2\le a_n\le e^{1/n}\log 2$.
Take the limit.
Recall:
MVT for Integrals
Let $g: [a,b]  \rightarrow \mathbb{R_{+}}$ be a non-negative integrable function. Then for every continuos function $f: [a,b] \rightarrow \mathbb{R}$ there is a $t \in [a,b]$ s.t.
$\displaystyle{\int_{a}^{b}}f(x)g(x)dx =f(t)\int_{a}^{b}g(x)dx.$
