Find $\lim_{n\to\infty} n^2\int_0^1 \frac{dx}{(1+x^2)^n }. $ 
Find $$\lim_{n\to\infty} n^2\int_0^1 \frac{dx}{(1+x^2)^n }. $$

Actually, note that $f_n(x)= 1/(1+x^2)^n$ is a pointwise convergent and uniformly bounded sequence of functions where the limit function is  $f(x)= 0$ if $x\neq 0$ and $f(0) = 1$.
Hence, by Arzela's theorem of interchanging of limit and integration, we have , $$\lim_{n\to\infty} \int_0^1 \frac{dx}{(1+x^2)^n } = \int_0^1  \lim_{n\to\infty} f_n(x) dx = 0, $$ hence the limit is of $\infty \cdot 0$ form , and I tried some inequalities but can't reach some decent step .
 A: $$\int_0^1{\mathrm{dx}\over(1+x^2)^n}\geq\int_0^{1/\sqrt{n}}{\mathrm{dx}\over(1+x^2)^n}\geq\int_0^{1/\sqrt{n}}{\mathrm{dx}\over\left(1+\frac1n\right)^n}\geq{1\over\sqrt{n}}{1\over2e}$$
for large $n$.  Therefore, the limit you seek is $\infty.$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
& \color{#44f}{\lim_{n\to\infty}\bracks{n^{2}\int_{0}^{1}{\dd x \over \pars{1 + x^{2}}^{n}}}} =
\lim_{n\to\infty}\bracks{n^{2}\int_{0}^{1}
\expo{-n\ln\pars{1 + x^{2}}}\,\,\,\dd x}
\\[5mm] = & \
\lim_{n\to\infty}\,\,\bracks{n^{2}\ \overbrace{\int_{0}^{\infty}
\expo{-nx^{2}}\,\,\,\dd x}^{\ds{Laplace's\ Method}}} =
\lim_{n\to\infty}\,\,\bracks{n^{2}\,{1 \over \root{n}}\int_{0}^{\infty}
\expo{-x^{2}}\,\,\,\dd x} = \bbx{\color{#44f}{\infty}}\\ &
\end{align}
