Binomial limit $\left(\binom{3n}{n}\binom{2n}{n}^{-1}\right)^{1/n}$ as $n\to \infty$

The limit:$$\lim_{n\to \infty}\left(\dfrac{\binom{3n}{n}}{\binom{2n}{n}}\right)^\frac{1}{n}$$

What I did was put limit = $L$. Then,

$$\begin{align}\log(L)&={\lim_{n \to \infty}}\dfrac{1}{n}\cdot\sum_{r=0}^{{n-1}} \log\left(\dfrac{3-\frac{r}{n}}{2-\frac{r}{n}}\right)\\ &=\int_0^1 \log\left(\dfrac{3-x}{2-x}\right)dx\\ &=\log\left(\dfrac{27}{16}\right) \end{align}$$

Is this aproach correct? Is there other method.

Edit: I have corrected the expression for the limit.

Solution 1:

$$ \begin{aligned} \lim _{n\to \infty }\left(\frac{\binom{3n}{n}}{\binom{2n}{n}}\right)^{\frac{1}{n}} & = \lim _{n\to \infty }\left(\frac{n!\left(3n\right)!}{\left(2n\right)!^2}\right)^{\frac{1}{n}} \\& = \lim _{n\to \infty }\exp\left(\frac{\ln\left(\frac{n!\left(3n\right)!}{\left(2n\right)!^2}\right)}{n}\right) \\& = \lim _{n\to \infty }\exp\left(\frac{\ln\left(\frac{\left(\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\right)\left(\sqrt{2\pi \left(3n\right)}\left(\frac{3n}{e}\right)^{3n}\right)}{\left(\sqrt{2\pi \left(2n\right)}\left(\frac{2n}{e}\right)^{2n}\right)^2}\right)}{n}\right) \\& = \lim _{n\to \infty }\exp\left(\frac{\ln\left(\frac{3^{\frac{6n+1}{2}}}{2^{4n+1}}\right)}{n}\right) \\& = \lim _{n\to \infty \:}\exp \left(\frac{\ln \left(\frac{e^{\left(\frac{6n+1}{2}\right)\ln3}}{e^{(4n+1)\ln2}}\right)}{n}\right) \\& = \lim _{n\to \infty }\exp \left(\ln \left(\frac{27}{16}\right)-\frac{\ln \left(\frac{4}{3}\right)}{2n}\right) \\& = \color{red}{\frac{27}{16}} \end{aligned} $$

Solved with Stirling approximation $$x! \approx \sqrt{2\pi x}\left(\frac{x}{e}\right)^x, \text{ for } x \to \infty$$