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.
 A: $$\lim_{n \rightarrow \infty} \frac{\ln(2n^n)-\ln(\sqrt n)}{\ln(3n^{2n}) + \ln(\sqrt n)} = \lim_{n \rightarrow \infty} \frac{n \ln{n} +  \ln{2} - \frac{1}{2} \ln{n}}{2 n \ln{n} +  \ln{3} + \frac{1}{2} \ln{n}} $$
Note that, in both the numerator and denominator, the $n \ln{n}$ terms dominate the others in this limit.  We can ignore the last two terms in each of the numerator and denominator in this limit, and the result is
$$\lim_{n \rightarrow \infty} \frac{\ln(2n^n)-\ln(\sqrt n)}{\ln(3n^{2n}) + \ln(\sqrt n)} = \lim_{n \rightarrow \infty} \frac{ n \ln{n}}{2 n \ln{n}} = \frac{1}{2} $$
To be more specific, we can factor the $n \ln{n}$ terms out:
$$ \lim_{n \rightarrow \infty} \frac{n \ln{n} +  \ln{2} - \frac{1}{2} \ln{n}}{2 n \ln{n} +  \ln{3} + \frac{1}{2} \ln{n}} = \lim_{n \rightarrow \infty} \frac{1+\frac{\ln{2}}{n \ln{n}} - \frac{1}{2 n}}{2+\frac{\ln{3}}{2 n\ln{n}} + \frac{1}{2 n}} $$
and see the result.
A: HINT: 
$$\begin{align*}\frac{\ln(2n^n)-\ln(\sqrt n)}{\ln(3n^{2n}) + \ln(\sqrt n)}&=\frac{\ln 2n^{n-\frac12}}{\ln 3n^{2n+\frac12}}\\\\
&=\frac{\ln 2+\left(n-\frac12\right)\ln n}{\ln 3+\left(2n+\frac12\right)\ln n}\\\\
&=\frac{\frac{2\ln 2}{\ln n}+2n-1}{\frac{2\ln 3}{\ln n}+4n+1}
\end{align*}$$
