Let $$f: [0,1] \times [0,1] \rightarrow R $$ a continuous function. Evaluate this limit:
This is what I've done: f(x) is defined on $$ [0,1] \times [0,1]$$ so f is limited $$m \le f(x,y) \le M$$ Now $$ m\lim_{n\to\infty} (\frac{(2n+1)!}{(n!)^2})^2 \iint_{0}^{1}(xy(1-x)(1-y))^ndxdy \le \lim_{n\to\infty} (\frac{(2n+1)!}{(n!)^2})^2 \iint_{0}^{1}(xy(1-x)(1-y))^nf(x,y)dxdy \le M \lim_{n\to\infty} (\frac{(2n+1)!}{(n!)^2})^2 \iint_{0}^{1}(xy(1-x)(1-y))^ndxdy $$
$$\iint_{0}^{1}xy(1-x)(1-y))^ndxdy=I^2$$ with $$I=\int_{0}^{1}x^n(1-x)^ndx= \frac{\Gamma(n+1) \Gamma(n+1)}{\Gamma(2n+2)}= \frac{(n!)^2}{(2n+1)!}$$
So $$ m \le \lim_{n\to\infty} (\frac{(2n+1)!}{(n!)^2})^2 \iint_{0}^{1}(xy(1-x)(1-y))^nf(x,y)dxdy \le M $$ but from this point I'm stuck
Using Stirling's approximation we can simplify the outer term asymptotically to be
$$\left(\frac{(2n+1)!}{(n!)^2}\right)^2 = (2n+1)^2{2n \choose n}^2 \sim \frac{(2n+1)^22^{4n}}{\pi n}$$
and denote
$$I_n \equiv \frac{(2n+1)^22^{4n}}{\pi n}\int_0^1\int_0^1(xy(1-x)(1-y))^nf(x,y)dxdy$$
With four factors of $2^n$ to conveniently distribute amongst our four factors raised to the $n$th power within the integrand, consider the variable interchange $(2x,2y) \leftrightarrow (x,y)$:
$$I_n = \frac{\left(n+\frac{1}{2}\right)^2}{\pi n}\int_0^2\int_0^2(xy(2-x)(2-y))^nf\left(\frac{x}{2},\frac{y}{2}\right)dxdy$$
Now break this domain into four squares with the lines $x=1$ and $y=1$ and consider the substitution
$$\begin{cases}\begin{cases}u = (2x-x^2)^{n+1} \\ v = (2y-y^2)^{n+1}\end{cases} \\ \\ \begin{cases}du = 2(n+1)(2x-x^2)^n|1-x|dx \\ dv = 2(n+1)(2y-y^2)^n|1-y|dy\end{cases}\end{cases} \implies \begin{cases}x = 1 \pm \sqrt{1-u^{\frac{1}{n+1}}} \\ y = 1 \pm \sqrt{1-v^{\frac{1}{n+1}}}\end{cases}$$
where the choice in $\pm$ depends on the square. This gives us the integral
$$I_n = \frac{1}{4\pi n}\left(\frac{n+\frac{1}{2}}{n+1}\right)^2\int_0^1\int_0^1\frac{\sum\limits_{\pm}f\left(\frac{x_\pm(u)}{2},\frac{y_\pm(v)}{2}\right)}{\sqrt{1-u^{\frac{1}{n+1}}}\sqrt{1-v^{\frac{1}{n+1}}}}dudv$$
where the sum $\sum\limits_{\pm}$ is taken over all permutations of $\pm$. Taking the limit we obtain
$$\lim_{n\to\infty}I_n = \lim_{n\to\infty} \frac{1}{4\pi}\left(\frac{n+\frac{1}{2}}{n+1}\right)^2\int_0^1\int_0^1\frac{\sum\limits_{\pm} f\left(\frac{x_\pm(u)}{2},\frac{y_\pm(v)}{2}\right)}{\sqrt{n-nu^{\frac{1}{n+1}}}\sqrt{n-nv^{\frac{1}{n+1}}}}dudv$$
$$\longrightarrow \frac{4f\left(\frac{1}{2},\frac{1}{2}\right)}{4\pi}\left[\int_0^1\frac{du}{\sqrt{-\log u}}\right]^2$$
The last integral, with a little extra work, is an inverse reciprocal Gaussian
$$\int_0^1\frac{du}{\sqrt{-\log u}} = \int_0^\infty 1 - e^{-\frac{1}{v^2}}dv = \sqrt{\pi}$$
which can be seen by geometrically looking at the graph $v = \frac{1}{\sqrt{-\log u}}$ and deciding that the inverse function is easier to integrate. This gives us a final answer of
$$\lim_{n\to\infty}\left(\frac{(2n+1)!}{(n!)^2}\right)^2\int_0^1\int_0^1(xy(1-x)(1-y))^nf(x,y)dxdy = \lim_{n\to\infty}I_n = \boxed{\frac{f\left(\frac{1}{2},\frac{1}{2}\right)}{\pi^2}}$$
by dominated convergence.