Dominated Convergence Theorem Exercise I am asked to find $$\lim_{n \to \infty} \int_0^\infty n^2e^{-nx} \tan^{-1} x \, dx.$$
Here is my attempt.
Write $$\int_0^\infty n^2e^{-nx}\tan^{-1}x \, dx=\int_0^1 n^2e^{-nx} \tan^{-1} x \,dx + \int_1^\infty n^2e^{-nx}\tan^{-1} x \, dx$$
$$=\int_0^{n^2} e^{-\frac x n} \tan^{-1}\left(\frac x {n^2}\right) \, dx+\int_1^\infty n^2 e^{-nx} \tan^{-1}x \, dx.$$
Then note that $$\left| 1_{(0,n^2)}(x)e^{-x/n}\tan^{-1} \left(\frac x {n^2}\right) \right| \le \frac \pi 2$$ for all $x>0$ and all $n\ge 1$
and $$|n^2e^{-nx}\tan^{-1}x| \le \frac{\pi}{2}\frac 2 {x^2}$$ for all $x\in [1,\infty)$ and all $n\ge 1$. Thus the dominated convergence gives $${\lim_{n\to\infty} \int_0^\infty 1_{(0,n^2)}(x)e^{-x/n}\tan^{-1} \left(\frac x {n^2}\right) \, dx = 0}$$ and $$\lim_{n\to\infty} \int_1^\infty n^2e^{-nx} \tan^{-1}x\,dx=0,$$ and hence $$\lim_{n \to \infty}\int_0^\infty n^2e^{-nx}\tan^{-1}x\,dx=0.$$
Is this correct?
EDIT: Unfortunately the above is not correct (see Dr. MV's comment). The correct justification is shown below (given by Sangchul Lee).
$$\int_0^\infty n^2e^{-nx} \tan^{-1} xdx=\int_0^\infty ne^{-x} \tan^{-1} (\frac{x}{n}) \, dx.$$ Since $$|ne^{-x}\tan^{-1} (\frac{x}{n})|\le xe^{-x}$$ for all $x>0$ and all $n\ge1$ we deduce that  $$\lim_{n \to \infty} \int_0^\infty n^2e^{-nx} \tan^{-1} x \, dx=\lim_{n\to\infty}\int_0^\infty ne^{-x} \tan^{-1} (\frac{x}{n}) \, dx=\int_0^\infty \lim_{n
\to\infty}ne^{-x} \tan^{-1} (\frac{x}{n}) \, dx=\int_0^\infty xe^{-x} \, dx=1.$$ The point is that $\tan^{-1}x\le x$ for all $x\ge0$, an inequality I had forgotten!
 A: 
I thought it might be instructive to present a way forward that circumvents the use of the Dominated Convergence Theorem.  To that end, we proceed.

Integrating by parts the integral of interest with $u=\arctan(x)$ and $v=-ne^{-nx}$ yields 
$$\begin{align}
\int_0^\infty n^2e^{-nx}\arctan(x)\,dx&=\int_0^\infty \frac{ne^{-nx}}{1+x^2}\,dx\\\\
&=\color{blue}{\int_0^\infty ne^{-nx}\,dx}-\color{red}{\int_0^\infty \frac{nx^2e^{-nx}}{1+x^2}\,dx}\\\\
&=\color{blue}{1}+\color{red}{O\left(\frac1{n^2}\right)}\tag 1
\end{align}$$
whence passing to the limit reveals the coveted result

$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\int_0^\infty n^2e^{-nx}\arctan(x)\,dx=1}$$


NOTE:
Here, we show that the $ \displaystyle \int_0^\infty \frac{nx^2e^{-nx}}{1+x^2}\,dx=O\left(\frac1{n^2}\right)$.  Enforcing the substitution $x\to x/n$ in $(1)$ yields
$$\begin{align}
\int_0^\infty \frac{nx^2e^{-nx}}{1+x^2}\,dx&=\frac1{n^2}\int_0^\infty \frac{x^2e^{-x}}{1+(x/n)^2}\,dx\\\\
&\le \frac1{n^2}\int_0^\infty x^2e^{-x}\,dx\\\\
&=\frac2{n^2}
\end{align}$$
as was to be shown.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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 \newcommand{\mc}[1]{\mathcal{#1}}
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 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\lim_{n \to \infty}\int_{0}^{\infty}n^{2}\expo{-nx}\arctan\pars{x}
\,\dd x:\ ?}$.

$$\bbox[#ffe,10px,border:1px dotted navy]{%
\mbox{Besides the 'original motivation', it can be evaluated as follows:}}
$$

\begin{align}
&\lim_{n \to \infty}\int_{0}^{\infty}n^{2}\expo{-nx}\arctan\pars{x}
\,\dd x
\\[5mm] = &\
\lim_{n \to \infty}\braces{%
n^{2}\int_{0}^{\infty}\expo{-nx}x\,\dd x +
n^{2}\int_{0}^{\infty}\expo{-nx}\bracks{\arctan\pars{x} - x}\dd x}
\end{align}

Moreover,
$\ds{\arctan\pars{x} - x = -\,{\xi^{2} \over \xi^{2} + 1}\,x\quad}$ for some $\ds{\quad\xi\ {\large\mid}\ 0 < \xi < x > 0}$ such that:
\begin{align}
& 0 < \verts{n^{2}\int_{0}^{\infty}\expo{-nx}\bracks{\arctan\pars{x} - x}\dd x} <
n^{2}\int_{0}^{\infty}\expo{-nx}x^{3}\,\dd x = {6 \over n^{2}}
\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\to}\,\,\, \color{#f00}{\large 0}
\\[5mm]
&\mbox{and}\quad
\lim_{n \to \infty}\pars{n^{2}\int_{0}^{\infty}\expo{-nx}x\,\dd x} =
\color{#f00}{\large 1}
\end{align}

$$
\implies\bbx{\ds{%
\lim_{n \to \infty}\int_{0}^{\infty}n^{2}\expo{-nx}\arctan\pars{x}\,\dd x = 1}}
$$
