$$\lim_{n\to \infty} \left(\frac{(2n)!}{n!n^n}\right)^{\frac{1}{n}}$$
Where I started:
For me, this looked like a $1^{\infty}$ form. So I did use the 'monkey-on-the-tree' kind of methodology, where we say,
'Get the monkey down and chop off the tree.'
That is, get the $\frac{1}{n}$ down, and impose the limit to the rest, subtracting one from it. It's a trick to solve these faster...
$${\frac{1}{n}} \lim_{n\to \infty} \left(\frac{(2n)!}{n!n^n}-1\right)$$
But now I am stuck. Can anyone please help me out?
Assuming that the limit exists, let's let $$\begin{align} L &= \lim_{n \to \infty} \left(\frac{(2n)!}{n! n^n} \right)^{1/n} \end{align}$$ Then taking the logarithm, $$\begin{align} \ln(L) &= \lim_{n \to \infty} \frac{\ln((2n)!) - \ln(n!) - n\ln(n)}{n} \end{align}$$ I assume this is what you mean by "Getting the monkey down".
The opening wedge here is Stirling's approximation: $$ \ln(n!) = n\ln(n) - n + O(\ln(n))$$ for large $n$. Then asymptotically, $$\begin{align} \frac{\ln((2n)!) - \ln(n!) - n\ln(n)}{n} &= \frac{\left[2n\ln(2n) - 2n \right] - \left[n\ln(n) - n \right] - n\ln(n) + O(\ln(n))}{n}\\ &= \frac{(2\ln(2) - 1)n}{n} + \frac{O(\ln(n))}{n}\\ \Rightarrow\ln(L) = \lim_{n \to \infty}\frac{\ln((2n)!) - \ln(n!) - n\ln(n)}{n} &= 2\ln(2) - 1\\ \Rightarrow L &= e^{2\ln(2) - 1} = \frac{4}{e} \end{align}$$