Calculate $\lim_{n\rightarrow\infty} \int_0^1 nx^2(1-x^2)^n \, dx$ I have to calculate $$\lim_{n\rightarrow\infty} \int_0^1 nx^2(1-x^2)^n \, dx$$
I've created the series $(f_n)_{n\in \mathbb{N}}$ with $f_n:[0,1]\rightarrow \mathbb{R}, f_n(x)=nx^2(1-x^2)^n$. I considered $x\in[0,1]$ a scalar and proceed to calculated the limit $\lim_{n\rightarrow\infty}f_n(x)$, which equals to $0, \forall x\in[0,1)$ (if I'm not wrong), but I'm getting stuck at calculating the limit for the particular case of $x=1$. Thanks for help!  
 A: Alternative approach: by setting $x=\sin\theta$ we have
$$ I(n) = \int_{0}^{1}nx^2(1-x^2)^n\,dx = n\int_{0}^{\pi/2}\sin^2(\theta)\cos^{2n+1}(\theta)\,d\theta \tag{1}$$
and since over the interval $\left[0,\frac{\pi}{2}\right]$ both $\sin(\theta)$ and $\cos(\theta)$ are non-negative, but $\sin(\theta)\leq\theta$ and $\cos(\theta)\leq e^{-\theta^2/2}$, it follows that:
$$ 0\leq I(n) \leq \int_{0}^{\pi/2} n\theta^2 e^{-\left(n+\frac{1}{2}\right)\theta^2}\,d\theta \leq  \int_{0}^{+\infty} n\theta^2 e^{-\left(n+\frac{1}{2}\right)\theta^2}\,d\theta=\sqrt{\frac{\pi}{2}}\frac{n}{(2n+1)^{3/2}}\tag{2}$$
and the wanted limit is zero by squeezing.
A: Since $1-x^2<e^{-x^2}$ for each $0\leq x\leq 1$, the integrand is non-negative and bounded above by $nx^2e^{-nx^2}$. Since $ye^{-y}$ is a bounded function on $[0,\infty)$, and $nx^2e^{-nx^2}\to 0$ for all $x\geq 0$, the result follows from the dominated convergence theorem.
A: Integrating by parts, 
$$ \int_0^1 nx^2(1-x^2)^n \, dx = \left[ -\frac{n}{2(n+1)}x(1-x^2)^{n+1} \right]_0^1 + \frac{n}{2(n+1)}\int_0^1 x(1-x^2)^{n+1} \, dx \\
= \frac{n}{2(n+1)}\int_0^1 (1-x^2)^{n+1} \, dx $$
The fraction converges to $1/2$, so now we need to look at the remaining integral. $(1-x^2)^{n+2} < (1-x^2)^{n+1} $, so the integrand decreases as $n \to \infty$. We can then use the monotone convergence theorem or similar to show the integral tends to zero, so the limit is zero.
A: 
The integral can be evaluated in closed form before letting $n\to \infty$.  To that end we now proceed.

Enforcing the substitution $x\to \sqrt{x}$, we find
$$\begin{align}
\int_0^1 nx^2(1-x^2)^n\,dx&=\frac n2\int_0^1 x^{1/2}(1-x)^n\,dx\\\\
&=\frac n2 B\left(\frac32,n+1\right)\\\\
&=\frac n2 \frac{\Gamma(3/2)\Gamma(n+1)}{\Gamma(n+5/2)}\\\\
&=\frac n2 \frac{\color{blue}{\Gamma(3/2)}\color{orange}{\Gamma(n+1)}}{(n+3/2)(n+1/2)\color{red}{\Gamma(n+1/2)}}\\\\
&=\frac n2 \frac{\color{blue}{\frac{\sqrt\pi}{2}}\color{orange}{(n!)}}{(n+3/2)(n+1/2)\color{red}{\left(\frac{2^{1-2n}\sqrt\pi\,\Gamma(2n)}{\Gamma(n)}\right)}}\\\\
\therefore \int_0^1 nx^2(1-x^2)^n\,dx&=\frac{4^n\,n\,(n!)^2}{(2n+3)(2n+1)(2n)!}\\\\
&=O\left(\frac{1}{\sqrt n}\right)\,\,\,\,\dots \text{Applying Stirling's Formula}
\end{align}$$
Hence, as $n\to \infty$, the integral of interest goes to $0$ as $\frac1{\sqrt n}$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\lim_{n \to \infty}\int_{0}^{1}nx^{2}\pars{1 -x^{2}}^n\,\dd x & =
\lim_{n \to \infty}\bracks{n%
\int_{0}^{1}\exp\pars{2\ln\pars{x} + n\ln\pars{1 - x^{2}}}\,\dd x}
\end{align}

The $\ds{\exp}$-argument has a 'sharp maximum' at
  $\ds{x_{n} = \pars{n + 1}^{-1/2}}$ such that

\begin{align}
\lim_{n \to \infty}\int_{0}^{1}nx^{2}\pars{1 -x^{2}}^n\,\dd x & =
\lim_{n \to \infty}\bracks{\pars{n \over n + 1}^{n + 1}%
\int_{0}^{\infty}\exp\pars{-\,{\bracks{x - x_{n}}^{\,2} \over 2\sigma_{n}^{2}}}\,\dd x}
\end{align}

where $\ds{\sigma_{n} \equiv {\root{n} \over 2\pars{n + 1}}}$.

Then,
\begin{align}
\lim_{n \to \infty}\int_{0}^{1}nx^{2}\pars{1 -x^{2}}^n\,\dd x & =
\expo{-1}\root{\pi \over 2}\lim_{n \to \infty}\braces{%
\sigma_{n}\bracks{1 + \mrm{erf}\pars{x_{n} \over \root{2}\sigma_{n}}}}
\\[5mm] & =
\expo{-1}\root{\pi \over 2}\bracks{1 + \mrm{erf}\pars{\root{2}}}
\lim_{n \to \infty}\sigma_{n} = \bbx{0}
\end{align}
