I am trying to solve:
$$\lim_{n\to\infty}n^\frac{\ln(n)}{n}$$
Here's what I did but I am not sure if it's right.
$$S=\lim_{n\to\infty}n^\frac{\ln(n)}{n}$$ $$\implies\ln(S)=\lim_{n\to\infty}\ln\left(n^\frac{\ln(n)}{n}\right)=\lim_{n\to\infty}\frac{\ln^2(n)}{n}$$
Applying L'Hôpital's rule:
$$\begin{align} &\implies\ln(S)=\lim_{n\to\infty}\frac{2\ln(n)}{n} \\ &\implies\ln(S)=\lim_{n\to\infty}\frac{2}{n}=0 \\ &\implies S=1 \\ \end{align}$$
Is this correct and is this an ideal approach? Are there any better alternative approaches to this limit?
Yes, your proof is fine. Here's another one just for fun, using the squeeze theorem.
The limit is clearly at least $1$. On the other hand, we have for $n$ sufficiently large (e.g. $n \ge 3$):
$$n^{\ln n / n} \le n^{\sqrt{n} / n} = n^{1/\sqrt{n}} = \left(\sqrt{n}^{1/\sqrt{n}}\right)^2.$$
Now using the well-known limit $\lim_{t \to \infty} t^{1/t} = 1$, the result follows.