TITLE: How find this limit $\lim_{n\to\infty}\frac{(2n+1)!}{(n!)^2}\int_{0}^{1}(x(1-x))^nf(x)dx$
QUESTION [6 upvotes]: Let $f:[0,1]\longrightarrow R$ be a continuous function,Calculate the limit
  $$\lim_{n\to\infty}\dfrac{(2n+1)!}{(n!)^2}\int_{0}^{1}(x(1-x))^nf(x)dx$$

My try:use this

$$n!\approx\left(\dfrac{n}{e}\right)^n\sqrt{2n\pi}$$
  so
  $$\dfrac{(2n+1)!}{(n!)^2}\approx\dfrac{\left(\dfrac{2n+1}{e}\right)^{2n+1}\sqrt{2(2n+1)\pi}}{\left(\dfrac{n}{e}\right)^{2n}\cdot 2n\pi}=\dfrac{\sqrt{2n+1}}{e\sqrt{2\pi}}\left(2+\dfrac{1}{n}\right)^{2n+1}$$

Then following I can't it,and  I guess this problem answer is $f(\dfrac{1}{2})$,But I can't prove it.
Thank you for your help.

REPLY [3 votes]: Write
$$
[x(1-x)]^n = \exp\Bigl\{n\log[x(1-x)]\Bigr\}
$$
and observe that $\log[x(1-x)]$ has a maximum at $x=1/2$.  Expanding this in a power series about this point yields
$$
\log[x(1-x)] = -\log 4 - 4\left(x-\frac{1}{2}\right)^2 + o\left(x-\frac{1}{2}\right)^2
$$
as $x \to 1/2$, so we can conclude that
$$
\begin{align}
\int_0^1 \exp\Bigl\{n\log[x(1-x)]\Bigr\}f(x)\,dx &\sim \int_{-\infty}^{\infty} \exp\left\{n\left[-\log 4 - 4\left(x-\frac{1}{2}\right)^2\right]\right\}f\left(\frac{1}{2}\right)\,dx \\
&= 2^{-2n-1} \sqrt{\frac{\pi}{n}} f\left(\frac{1}{2}\right)
\end{align}
$$
by the Laplace method.  Now
$$
\frac{(2n+1)!}{(n!)^2} \sim 2^{2n+1} \sqrt{\frac{n}{\pi}}
$$
by Stirling's formula, from which the result follows.