Tried this question here How to calculate $\lim\limits_{n \to \infty} \left( \frac{\ln(n+1)^{n+1}}{\ln n^n} \right)^n$? and was curious about the result. The answer according to Wolfram Alpha is $e$, so I wanted to try it.
$\lim\limits_{n \to \infty} \left( \frac{\ln((n+1)^{n+1})}{\ln (n^n)} \right)^n$
$\lim\limits_{n \to \infty} \left( \frac{(n+1)\ln(n+1)}{n\ln (n)} \right)^n$
$\lim\limits_{n \to \infty} \left( \frac{\ln(n+1)}{\ln(n)}\left(1 + \frac{1}{n}\right) \right)^n$
This is similar to the typical definition $\lim\limits_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e$ but it has the extra log factors.
How come these two happen to be equivalent? Is it valid to apply L'Hospital's Rule to the logs even though they're inside the $()^n$? Or can it be applied to just part of the function and not the other half? What's the correct way to handle this extra log multiplier?
For instance:
$\lim\limits_{n \to \infty}\frac{\ln(n+1)}{\ln(n)} = \lim\limits_{n \to \infty}\frac{\frac{d}{dn}\ln(n+1)}{\frac{d}{dn}\ln(n)} = \lim\limits_{n \to \infty}\frac{n}{1+n} = \lim\limits_{n \to \infty}\frac{1}{1/n+1} = 1$
but I don't think we can necessarily analyze this "separately" from the main result; I think they must be taken together somehow. I also considered squeeze theorem but couldn't think of another function approaching $e$ from the other side.