Find $\lim_{n \to \infty} \left( \frac{3^{3n}(n!)^3}{(3n)!}\right)^{1/n}$ Find $$\lim_{n \to \infty} \left( \frac{3^{3n}(n!)^3}{(3n)!}\right)^{1/n}$$
I don't know what method to use, if  we divide numerator and denominator with $3^{3n}$, I don't see that we win something. I can't find two sequences, to use than the squeeze theorem. I'm stuck.

Use equivalents  and Stirling's formula: $\;n!\sim_\infty \sqrt{2\pi n}\Bigl(\dfrac n{\mathrm e}\Bigr)^{\!n}$:
$$\biggl(\!\frac{3^{3n}(n!)^3}{(3n)!}\!\biggr)^{1/n}\!\sim_\infty\left(\frac{3^{3n}\sqrt{(2\pi n)^3\strut}\Bigl(\dfrac{n}{\mathrm e}\Bigr)^{\!3n}}{\sqrt{6\pi n}\Bigl(\dfrac{3n}{\mathrm e}\Bigr)^{\!3n}}\right)^{\!\tfrac 1n}=\Bigl(\sqrt{\tfrac 43} \pi n\Bigr)^{\!\tfrac 1n}\to 1.$$
