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.