How find this limit $\lim_{n\to\infty}\frac{(2n+1)!}{(n!)^2}\int_{0}^{1}(x(1-x))^nf(x)dx$ 
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.
 A: Hint. If we put
$$ K_{n}(x) = \frac{(2n+1)!}{(n!)^{2}} x^{n}(1-x)^{n}, $$
then the total mass $\int_{0}^{1} K_{n}(x) \, dx$ equals $1$ for any $n$. (Just apply integration by parts $n$ times!) Show that this sequence of functions is an approximation to the identity by checking that for any $\delta > 0$ we have
$$\lim_{n\to\infty} \int\limits_{\delta \leq |x - 1/2| \leq 1/2} K_{n}(x) \, dx = 0.$$
You may find the Stirling's formula useful. Once this is proved, you can check that
$$ \int_{0}^{1} f(x) K_{n}(x) \, dx - f\left(\tfrac{1}{2}\right)
= \int_{0}^{1} \left\{ f(x) - f\left( \tfrac{1}{2}\right) \right\} K_{n}(x) \, dx $$
goes to zero as $n \to \infty$.
A: I haven't worked out all the details, but your guess should be correct. Here's some intuition.
The sequence of functions $\dfrac{(2n+1)!}{(n!)^2} (x(1-x))^n$ are an 'approximation to the identity'. As $n$ increases, the function becomes very peaked near $x=1/2$, and is nearly $0$ everywhere else. Furthermore, you should be able to compute 
$$\dfrac{(2n+1)!}{(n!)^2}\int_0^1 (x(1-x))^n \,dx = 1$$
 for all $n$. (How to do this, I'm not precisely sure, but you might try the substitution $y = x-1/2$ or continue w/ Stirling's formula).
Then, use the continuity of $f$ to find a small neighborhood of $x=1/2$ so that the values of $f$ are close to $f(1/2)$ in this neighborhood.
Then,
$$
\dfrac{(2n+1)!}{(n!)^2}\int_0^1 (x(1-x))^n f(x) \,dx - f(1/2) \\
= \dfrac{(2n+1)!}{(n!)^2}\int_0^1 (x(1-x))^n f(x) \,dx - f(1/2)\dfrac{(2n+1)!}{(n!)^2}\int_0^1 (x(1-x))^n  \,dx
$$
since this integral equals $1$, so it remains to show
$$
\left|\dfrac{(2n+1)!}{(n!)^2}\int_0^1 (x(1-x))^n \left(f(x) - f(1/2)\right) \,dx \right|
$$
is arbitarily small as $n\to\infty$.
To do this, use the neighborhood previously mentioned to estimate this integral; inside this neighborhood, $f(x)$ is close to $f(1/2)$, and outside this neighborhood, $\dfrac{(2n+1)!}{(n!)^2} (x(1-x))^n$ is very small.
A: 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.
