This problem was posted half a year ago by Pierre Mounir on a Facebook group and until now it received no answers. Since most of his problems that I saw were amazing I can bet this one it's worth the time. Wolfram returns the answer to be $2$, which is quite elegant for it's look.
I remembered about it yesterday and gave a try again taking for simplicity $k=1$ (I had no chance with a bigger number). Also my whole idea was to somehow get to a point where I can use $\lim\limits_{f\to 0}\frac{a^f-1}{f}=\ln a$, thus I started as: $$\lim_{n\to \infty} {\frac{5^\frac{n!}{(2n)!}-4^\frac{1}{n!}}{3^\frac{n!}{(2n)!}-2^\frac{1}{n!}}}=\lim_{n\to \infty} \left(\frac{4}{2}\right)^{\frac{1}{n!}}\left(\frac{5^\frac{n!}{(2n)!}}{4^{\frac1{n!}}}-1\right)\left(\frac{3^\frac{n!}{(2n)!}}{2^{\frac1{n!}}}-1\right)^{-1}$$ $$=\lim_{n\to \infty} \underbrace{\sqrt[n!]{2}}_{\to 1}\left(\sqrt[n!]{\frac{5^\frac{1}{(2n)!}}{4}}-1\right)\left(\sqrt[n!]{\frac{3^\frac{1}{(2n)!}}{2}}-1\right)^{-1}$$ Well, yes $\frac{5^\frac{1}{(2n)!}}{4}$ and the other one in the denominator equals to $1$, but still I don't see how to use that limit. Also I tried to take a logarithm on both sides or to use L'hospital, but looks like a dead end.
I would love if someone can spot the trick for solving this limit and land some help.
This is an expansion of my comment into an answer.
If the expression under limit is of the form $(A-B) /(C-D) $ then all of $A, B, C, D$ tend to $1$ and we can write $$A-B=B\cdot\frac{\exp(\log A-\log B) - 1}{\log A - \log B} \cdot(\log A - \log B) $$ The first two factors tend to $1$ and hence the numerator can be replaced by $\log A-\log B$. Proceeding in similar manner we see that the expression under limit can be simplified greatly as each term gets replaced by its logarithm.
The resulting expression is $$\frac{a\log 5-2b\log 2}{a\log 3 - b\log 2}$$ where $a=(n!) ^{k} /(2kn)!,b=(n!)^{-k}$ and clearly $a/b=(n!) ^{2k}/(2kn)!$ tends to $0$ so that the desired limit is $2$.
The limiting behavior of $a_n=(n!) ^{2 k} /(2kn)!$ can be concluded via ratio test. We have $$\frac{a_{n+1}}{a_n} =\frac{(n+1)^{2k}}{(2kn+1)\cdots(2kn+2k)}\to\frac{1}{(2k)^{2k}}<1$$ and hence $a_n\to 0$.