Computing $\lim_{n \rightarrow \infty} n^2 \int_0^{2n} e^{-n \vert x-n \vert} \log \left[ 1+ \frac{1}{x+1} \right]dx$ I want to show that 
$$\lim_{n \rightarrow \infty} n^2 \int_0^{2n} e^{-n \vert x-n \vert} \log \left[ 1+ \frac{1}{x+1} \right]dx=2.$$
The reason I think this is the limit is because if we make the substitution $y = n(x-n)$, then we obtain
$$ \int_{-n^2}^{n^2} e^{-\vert y \vert} \log \left[ \left( 1 + \frac{n}{n^2+n+y} \right)^n \right] dx. $$
Then if I can show $ \lim_{n \rightarrow \infty} \left( 1 + \frac{n}{n^2+n+y} \right)^n = e$, dominated convergence should give the rest. Certainly we have
$$\limsup_{n \rightarrow \infty} \left( 1 + \frac{n}{n^2+n+y} \right)^n \leq e$$
Is there a way to get the reverse inequality for the $\liminf$?
 A: $$
\begin{align}
&n^2\int_0^{2n}e^{-n|x-n|}\log\left(1+\frac1{x+1}\right)\,\mathrm{d}x\\
&=n^2\int_{-n}^ne^{-n|x|}\log\left(1+\frac1{x+n+1}\right)\,\mathrm{d}x\tag{1}\\
&=n\int_{-n^2}^{n^2}e^{-|x|}\log\left(1+\frac1{x/n+n+1}\right)\,\mathrm{d}x\tag{2}\\
&=n\int_{-n}^{n^2}e^{-|x|}\log\left(1+\frac1{x/n+n+1}\right)\,\mathrm{d}x\tag{3a}\\
&+n\int_{-n^2}^{-n}e^{-|x|}\log\left(1+\frac1{x/n+n+1}\right)\,\mathrm{d}x\tag{3b}\\
\end{align}
$$
Explanation:
$(1)$: substitute $x\mapsto x+n$
$(2)$: substitute $x\mapsto x/n$
$(3)$: split integral at $x=-n$
$\hphantom{(3)\text{:}}$ in $\text{(3a)}$, $n\log\left(1+\frac1{x/n+n+1}\right)\le \log\left(1+\frac1n\right)^n\le\log(e)=1$
$\hphantom{(3)\text{:}}$ in $\text{(3b)}$, $n\log\left(1+\frac1{x/n+n+1}\right)\le n\log(2)$ and $\int_{-n^2}^{-n}e^{-|x|}\,\mathrm{d}x\le e^{-n}$

Thus, by Dominated Convergence, taking the limit of $\text{(3a)}$,
$$
\begin{align}
\lim_{n\to\infty}n\int_{-n}^{n^2}e^{-|x|}\log\left(1+\frac1{x/n+n+1}\right)\,\mathrm{d}x
&=\int_{-\infty}^\infty e^{-|x|}\,\mathrm{d}x\\[6pt]
&=2\tag{4}
\end{align}
$$
since $\lim\limits_{n\to\infty}\left(1+\frac1{x/n+n+1}\right)^n=e$.
Furthermore, by the estimates given in the Explanation above,
$$
\begin{align}
\lim_{n\to\infty}n\int_{-n^2}^{-n}e^{-|x|}\log\left(1+\frac1{x/n+n+1}\right)\,\mathrm{d}x
&\le\lim_{n\to\infty}ne^{-n}\log(2)\\
&=0\tag{5}
\end{align}
$$
Adding $(4)$ and $(5)$, we get
$$
\bbox[5px,border:2px solid #C0A000]{\lim_{n\to\infty}n^2\int_0^{2n}e^{-n|x-n|}\log\left(1+\frac1{x+1}\right)\,\mathrm{d}x=2}\tag{6}
$$
A: To answer your last question :
$$n\ln\left(\frac{n}{n^2+n+y}\right)=n\frac{n}{n^2+n+y}+O\left(n^{-1}\right)$$
when $n\gg 1$, and
$$\lim_{n\to+\infty}\frac{n^2}{n^2+n+y}=1.$$
Now take the exponential to obtain $\mathrm{e}$ as the limit.
