Having the sequence $(a_{n})_{n\geq1}$, $a_{n}=\int_{0}^{1} x^{n}(1-x)^{n}dx$, find $\lim\limits_{n{\rightarrow}\infty} \frac{a_{n+1}}{a_{n}}$ Having the sequence $(a_{n})_{n\geq1}$, $a_{n}=\int_{0}^{1} x^{n}(1-x)^{n}dx$, find $\lim\limits_{n{\rightarrow}\infty} \frac{a_{n+1}}{a_{n}}$. Is this limit equal to $\lim\limits_{n{\rightarrow}\infty}(a_{n})^{\frac{1}{2}}$? I am unsure if I can apply l'Hospital to this limit, as the integral is a real value. I tried applying integration by parts but I ended nowhere.
Later edit: I am so sorry, it is $(1-x)^{n}$, not $(1-x^{n})$, I did this mistake out of tiredness.

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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}\,}\,}
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\begin{align}
&\bbox[10px,#ffd]{\int_{0}^{1}x^{n}\pars{1 - x}^{n}\dd x} =
\int_{-1/2}^{1/2}\pars{{1 \over 4} - x^{2}}^{n}\dd x =
{1 \over 2^{2n}}\int_{0}^{1}\pars{1 - x^{2}}^{n}\dd x
\\[5mm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,&
{1 \over 2^{2n}}\int_{0}^{\infty}\expo{-nx^{2}}\dd x =
{\root{\pi} \over 2^{2n + 1}\root{n}}
\end{align}
such that
$$
{a_{n + 1} \over a_{n}}
\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,
{1/4 \over \root{1 + 1/n}}
\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\to}\,\,\,\bbx{1 \over 4}
$$
