Find $\lim_{n\to\infty}\left(\frac{n!}{n^n}\right)^{1/n}$ I am referring to this
video on youtube which shows the determination of
$$
\lim_{n\to\infty}\left(\frac{n!}{n^n}\right)^{1/n}.
$$
In the video, the trick is to set
$$
L=\lim_{n\to\infty}\left(\frac{n!}{n^n}\right)^{1/n}
$$
and then to consider
$$
\ln(L)=\ln\left(\lim_{n\to\infty}\left(\frac{n!}{n^n}\right)^{1/n}\right).
$$
I have some problems with the following step: It is said that, since $\ln$ is a continuous function, one has
$$
\ln\left(\lim_{n\to\infty}\left(\frac{n!}{n^n}\right)^{1/n}\right)=\lim_{n\to\infty}\ln\left(\left(\frac{n!}{n^n}\right)^{1/n}\right)
$$
Is this really true? Doesn't this step only hold if the limit exists (what is exactly the task to show)?

A better way to understand what's going on is to state the connection as follows: For any sequence of positive numbers $a_n$ and any positive real number $L$, we have
$$\lim_{n\to\infty}a_n=L\iff\lim_{n\to\infty}\ln(a_n)=\ln(L)$$
If it's possible to evaluate the limit of the $\ln(a_n)$'s (as turns out to be the case here), then one simply takes $e$ to that limit to get $L$, which the equivalence tells you must be the (unique) limit of the $a_n$'s. (As an aside, the equivalence can be extended to allow for $L=0$ and $L=+\infty$ with appropriate understandings of $\ln(L)$ for those cases.)
