On an evaluation of a limit I want to evaluate
$$\lim_{n \to \infty} n^{3/2}\int_0^1 \frac{x^2}{(1+x^2)^n}\ dx$$
All that I needed is an intergrable control function $g(\cdot)$ independent of $n \in \mathbb{N}$ such that $n^{3/2} \frac{x^2}{(1+x^2)^n}\leq g(x)$, but I do not find direct control function anyway....
 A: By the change variable $x^2=\frac{u}{n} $ we find
$$n^{3/2}\int_0^1 \frac{x^2}{(1+x^2)^n}\ dx=\frac{1}{2}\int_0^n\frac{\sqrt{u}}{(1+\frac{u}{n})^n}du\to\frac{1}{2}\Gamma(\frac{3}{2})=\frac{\sqrt{\pi}}{4}$$
A: I would attack this a different way.  Rewrite the integral as
$$I(n) = \int_0^1 dx \, x^2 \, e^{-n \log{(1+x^2)}}$$
Note that the maximum value of the integrand is at $x_0=1/\sqrt{n-1}$.  Thus, as $n\to\infty$, the integral value is dominated by $x\in [x_0-\epsilon,x_0+\epsilon]$, for small $\epsilon$. Note also that $0<x_0 < 1$ for large $n$.  Thus we may write  
$$I(n) \sim \int_{x_0-\epsilon}^{x_0+\epsilon} dx \, x^2 \, e^{-n \log{(1+x^2)}}$$
with exponentially small error.  Further, because for large $n$, the values of $x$ in the integral are small, we may Taylor expand the log term to get, with again exponetially small error,
$$I(n) \sim \int_{x_0-\epsilon}^{x_0+\epsilon} dx \, x^2 \, e^{-n x^2}$$
And with still further small exponential error, we may extend the integration interval to $[0,\infty]$ to get
$$I(n) \sim \int_{0}^{\infty} dx \, x^2 \, e^{-n x^2} = \frac14 \sqrt{\frac{\pi}{n^3}}$$
The limit sought is therefore $\sqrt{\pi}/4$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\lim_{n \to \infty}\bracks{n^{3/2}
\int_{0}^{1}{x^{2} \over \pars{1 + x^{2}}^{n}}\,\dd x}}
\\[5mm] = &\
\lim_{n \to \infty}\bracks{%
n^{3/2}\int_{0}^{1}{\dd x \over \pars{1 + x^{2}}^{n - 1}} -
n^{3/2}\int_{0}^{1}{\dd x \over \pars{1 + x^{2}}^{n}}\,\dd x}
\\[5mm] = &\
\lim_{n \to \infty}\bracks{%
n^{3/2}\int_{0}^{1}\expo{-\pars{n - 1}\ln\pars{1 + x^{2}}}\,\,\,\dd x -
n^{3/2}\int_{0}^{1}\expo{-n\ln\pars{1 + x^{2}}}\,\,\,\dd x}
\\[5mm] = &\
\lim_{n \to \infty}\bracks{%
n^{3/2}\int_{0}^{\infty}\expo{-\pars{n - 1}x^{2}}\,\,\,\dd x -
n^{3/2}\int_{0}^{\infty}\expo{-nx^{2}}\,\,\,\dd x}\
\pars{\substack{\ds{Laplace's}\\[0.5mm] \ds{Method}}}
\\[5mm] = &\
\lim_{n \to \infty}\pars{%
n^{3/2}\,{\root{\pi} \over 2}{1 \over \root{n - 1}} -
n^{3/2}\,{\root{\pi} \over 2}{1 \over \root{n}}}
\\[5mm] = &\
{\root{\pi} \over 2}\lim_{n \to \infty}\pars{%
n^{3/2}\,{\root{n} - \root{n - 1} \over \root{n - 1}\root{n}}}
\\[5mm] = &\
{\root{\pi} \over 2}\lim_{n \to \infty}\bracks{%
{n^{3/2} \over
\root{n - 1}\root{n}\pars{\root{n} + \root{n - 1}}}}
\\[5mm] = &\
{\root{\pi} \over 2}\lim_{n \to \infty}\bracks{%
{1 \over
\root{1 - 1/n}\pars{1 + \root{1 - 1/n}}}}
\\[5mm] = &\
\bbx{\root{\pi} \over 4} \approx 0.4431 \\ &
\end{align}
