The Sky is the Limit.

The expression can be simplified to:- $\displaystyle \lim_{n\to\infty} \left(\frac{(3n)!}{n!}\right)^{\frac{1}{n}}\cdot \frac{1}{n^{2}}$

Using Stirlings Approximation:- We can approximate $\displaystyle n! \approx \left(\frac{n}{e}\right)^{n}\cdot \sqrt{2n\pi}$ .

Using this we can simplify the limit as.

$\displaystyle \lim_{n\to\infty}\frac{\frac{27n^{3}}{e^{3}}}{\frac{n}{e}}\cdot \frac{1}{n^2}=\frac{27}{e^{2}}$ .

Since there are $2n$ terms in the numerator, we can split the $n^{2n}$ between them:

$\begin{aligned}L&=\lim_{n\to\infty}\left(\frac{n+1}{n}\times\frac{n+2}{n}\times\frac{n+3}{n}\times\dots\times\frac{3n}{n}\right)^{\large\frac{1}{n}}\end{aligned}$

Now, let us consider $\ln L$ :

$\begin{aligned}\ln L&=\lim_{n\to\infty}\ln\left(\frac{n+1}{n}\times\frac{n+2}{n}\times\frac{n+3}{n}\times\dots\times\frac{3n}{n}\right)^{\large\frac{1}{n}}\\ &=\lim_{n\to\infty}\frac{1}{n}\ln\left(\frac{n+1}{n}\times\frac{n+2}{n}\times\frac{n+3}{n}\times\dots\times\frac{3n}{n}\right)\\ &=\lim_{n\to\infty}\frac{1}{n}\left[\ln\left(\frac{n+1}{n}\right)+\ln\left(\frac{n+2}{n}\right)+\ln\left(\frac{n+3}{n}\right)+\dots+\ln\left(\frac{3n}{n}\right)\right]\\ &=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{2n}\ln\left(\frac{n+k}{n}\right)\\ &=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{2n}\ln\left(1+\frac{k}{n}\right)\end{aligned}$

This is a Riemann sum, which we can turn into an integral:

$\begin{aligned}\ln L&=\int_0^2\ln(1+x)dx\\ &=\big[x\ln(1+x)\big]_0^2-\int_0^2\frac{x}{1+x}dx\\ &=2\ln(1+2)-0\ln(1+0)+\int_0^2 \frac{1}{1+x}-1dx\\ &=2\ln3+\big[\ln(1+x)-x\big]_0^2\\ &=2\ln3+\ln(1+2)-2-\ln(1+0)+0\\ &=2\ln3+\ln3-2-\ln1\\ &=3\ln3-2\\ \therefore L&=e^{3\ln3-2}\\ &=\left(e^{\ln3}\right)^3e^{-2}\\ &=\frac{3^3}{e^2}\\ &=\boxed{\large{\frac{27}{e^2}}}\end{aligned}$