# 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.

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$\ln ab=\ln a+\ln b$ and $\ln a^r=r\ln a$ –  L. F. Jan 29 '13 at 21:59
Indeed. And then identify the dominant term in numerator and denominator. –  Harald Hanche-Olsen Jan 29 '13 at 22:02

$$\lim_{n \rightarrow \infty} \frac{\ln(2n^n)-\ln(\sqrt n)}{\ln(3n^{2n}) + \ln(\sqrt n)} = \lim_{n \rightarrow \infty} \frac{n \ln{n} + \ln{2} - \frac{1}{2} \ln{n}}{2 n \ln{n} + \ln{3} + \frac{1}{2} \ln{n}}$$

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.

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Why the downvote? –  Ron Gordon Jan 29 '13 at 22:26
@Andre: Oy, I see. Thanks for pointing that out, it is now fixed. –  Ron Gordon Jan 29 '13 at 22:48
Thank you very much, chose this answer because of the good explanation. –  Dominik Winter Jan 29 '13 at 22:49
@DominikWinter: You're welcome. –  Ron Gordon Jan 29 '13 at 23:22
@AndréNicolas: many thanks again. I apologize for the sloppiness. –  Ron Gordon Jan 29 '13 at 23:23

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*}

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