Limit of logarithms without l'Hospital This is my first post so I hope you forgive any formatting mistakes. This is a task out of a training exam, I may add that we have not yet introduced l'Hospital or derivatives. We have to determine the following limit:
$$\lim_{n \rightarrow \infty} \frac{\ln(2n^n)-\ln(\sqrt n)}{\ln(3n^{2n}) + \ln(\sqrt n)}$$
I am stuck on this for quite some time now, I tried to apply the ln sum rules, but i can not find a way to solve this one.
I'm thankful for advice.

HINT: 
$$\begin{align*}\frac{\ln(2n^n)-\ln(\sqrt n)}{\ln(3n^{2n}) + \ln(\sqrt n)}&=\frac{\ln 2n^{n-\frac12}}{\ln 3n^{2n+\frac12}}\\\\
&=\frac{\ln 2+\left(n-\frac12\right)\ln n}{\ln 3+\left(2n+\frac12\right)\ln n}\\\\
&=\frac{\frac{2\ln 2}{\ln n}+2n-1}{\frac{2\ln 3}{\ln n}+4n+1}
\end{align*}$$
