Solve the limit $\lim _{x\to \infty \:}\left(\frac{\left(\left(2x\right)!\right)}{x^xx!}\right)^{\frac{1}{x}}$ We need to solve this limit $\lim _{x\to \infty \:}\left(\frac{\left(\left(2x\right)!\right)}{x^xx!}\right)^{\frac{1}{x}}$
 we guess that the answer is 1, as n approach infinity 1/n => 0 so anything besides 0 to the power of 0 = 1 (except 0 of course).
We need to ensure if we had the right answer, please write some hints here.
 A: $$
\lim _{x\to \infty \:}\left(\frac{\left(\left(2x\right)!\right)}{x^xx!}\right)^{\frac{1}{x}}
$$
Using Stirling's approximation
$$
(2x)!
\sim (2x/e)^{2x}\sqrt{4\pi x}$$
and
$$
x!\sim (x/e)^x\sqrt{2\pi x}
$$
therefore the ratio $$\frac{(2x)!}{x!}\sim 2^{2 x+\frac{1}{2}} e^{-x} x^x$$
and dividing by $x^x$, we have to compute the limit
$$
\lim_{x\to\infty}(2^{2 x+\frac{1}{2}} e^{-x})^{1/x}=\frac{1}{e}\lim_{x\to\infty}(2^{2 x+\frac{1}{2}} )^{1/x}=\frac{1}{e}\lim_{x\to\infty}e^{\frac{1}{x}(2x+1/2)\log 2}=\frac{e^{\log 4}}{e}=\frac{4}{e}\ .
$$
A: Here we rely only on Riemann sums to evaluate the limit.  We can write 
$$\lim _{x\to \infty }\left(\frac{\left(2x\right)!}{x^xx!}\right)^{\frac{1}{x}}
=\lim_{n\to \infty}e^{\frac1n \log((2n)!-\frac1n \log(n!)-\log(n)}$$
by continuity of the Gamma function.  
Then, note that we have
$$\begin{align}
\frac1n \log((2n)!-\frac1n \log(n!)-\log(n)&=\frac1n \sum_{k=n+1}^{2n}\log(k/n)\\\\
&=\frac1n \sum_{k=1}^n \log(1+k/n)\\\\
&\to \int_0^1 \log(1+x)\,dx\,\,\text{as}\,\,n\to \infty\\\\
&=\log(4/e)
\end{align}$$
Therefore, we arrive at the result 
$$\bbox[5px,border:2px solid #C0A000]{\lim _{x\to \infty }\left(\frac{\left(2x\right)!}{x^xx!}\right)^{\frac{1}{x}}=4e^{-1}}$$
A: Yet another idea (see Baby Rudin 3.37):
$$
\lim_{x\to\infty}\left(\frac{\left(\left(2x\right)!\right)}{x^xx!}\right)^{\frac{1}{x}} =
\lim_{x\to\infty}\frac{\frac{(2(x+1))!}{(x+1)^{(x+1)}(x+1)!}}{\frac{(2x)!}{x^x x!}} = 
\lim_{x\to\infty}\frac{(2x+2)(2x+1)x^x}{(x+1)^{x+1}(x+1)} = 
$$
$$
\lim_{x\to\infty}\frac{2(2x+1)}{(x+1)}
\lim_{x\to\infty}\left(\frac{x}{x+1}\right)^x = \frac4e 
$$
A: Note that
$$
\frac{(2x)!}{x^x x!}=\frac{2x\cdot(2x-1)\cdots(x+2)\cdot(x+1)}{x^x}=2\cdot\left(2-\frac{1}{x}\right)\cdots\left(1+\frac{2}{x}\right)\cdot\left(1+\frac{1}{x}\right)\\=\prod_{k=1}^{x}\left(1+\frac{k}{x}\right).
$$
So
$$
\begin{eqnarray}
\lim_{x\rightarrow\infty}\left(\frac{(2x)!}{x^x x!}\right)^{1/x}&=&\lim_{x\rightarrow\infty}\exp\left(\frac{1}{x}\sum_{k=1}^{x}\log\left(1+\frac{k}{x}\right)\right)\\&=&\exp\left(\lim_{x\rightarrow\infty}\frac{1}{x}\sum_{k=1}^{x}\log\left(1+\frac{k}{x}\right)\right)\\&=&\exp\left(\int_{0}^{1}\log(1+u)du\right)\\&=&\exp\left(\log4 - 1\right)\\&=&\frac{4}{e}.
\end{eqnarray}$$
