$\underset{n\rightarrow\infty}{\lim} \frac{f_n^{(n)}(\frac{1}{n})}{n!}$ I just try to solve a problem, but I'm not sure it's right or not.
Here is the problem:
Suppose $f_n(x)=x^n\ln{x}$,  $ n \in \mathbb{N}$.
Find
$$
 \lim_{n\to\infty} \frac{f_n^{(n)}(\frac{1}{n})}{n!}
$$
My idea is by Leibniz product rule
$$
\left[\left(\frac{1}{n}\right)^n\ln{\frac{1}{n}}\right]^{(n)}=\overset{n}{\underset{k=0}{\sum}}\binom{n}{k}\cdot k!\cdot n^{-2n-k}
$$
Since n approaches to infinity, I check the n-th term, $n!\cdot n^{-n}$.
With the denominator, $$ \lim_{n\to \infty}n^{-n}= e^{\lim_{n\to \infty}(-n\ln{n})}=0$$
I appreciate your feedback.

Since, $$f_n'(x)=nx^{n-1}\ln x +x^{n-1}$$ we get:
$$f_n^{(n)}(x)=n f_{n-1}^{(n-1)}(x)+(n-1)!$$
So if $a_n(x)=f_n^{(n)}(x)$ then:
$$a_n(x)=na_{n-1}(x)+(n-1)!$$
$$a_n(x)=n(n-1)a_{n-2}(x)+n(n-2)!+(n-1)!$$
Repeating:
$$a_n(x)=n!\log x+\sum_{k=1}^{n} \frac{n!}{k}$$
So:
$$\frac{a_n(1/n)}{n!}=-\log (n) +\sum_{k=1}^n\frac1k,$$
which converges to $\gamma,$ the Euler–Mascheroni constant.
