The limit $$\lim_{n\rightarrow\infty}\dfrac{(4n)^{4n}n^n}{(3n)^{3n}(2n)^{2n}}$$ can be calculated as $\lim_{n\rightarrow\infty}\dfrac{4^{4n}}{3^{3n}2^{2n}}=\lim_{n\rightarrow\infty}\left(\dfrac{4^4}{3^32^2}\right)^n=\infty$.
What about $$\lim_{n\rightarrow\infty}\dfrac{(4n-100)^{4n-100}n^n}{(3n)^{3n}(2n)^{2n}}?$$ Would it still be $\infty$?
Call your function $f(n)$. Then it's enough to show that $\lim_{n \to \infty} \log f(n) = \infty$. We can write
$\log f(n) = (4n-100) \log (4n-100) + n \log n - 2n \log 2n - 3n \log 3n$
and let's consider that first term, $(4n-100) \log (4n-100)$. It would be really nice if that were $4n \log 4n$, so let's rewrite it as
$(4n-100) \log (4n-100) = (4n-100) \log 4n + (4n-100) \log ((4n-100)/4n)$
or, after some rearrangement,
$4n \log 4n - 100 \log 4n + (4n-100) \log (1 - 25/n)$.
Therefore the original function is
$\log f(n) = (4n \log 4n + n \log n - 2n \log 2n - 3n \log 3n) - 100 \log 4n + (4n-100) \log (1 - 25/n)$.
Now, let's start rearranging, recalling that $\log 4n = \log 4 + \log n$ and so on. You get
$\log f(n) = 4n \log 4 + 4n \log n + n \log n - 2n \log 2 - 2n \log n - 3n \log 3 - 3n \log n - 100 \log 4n + (4n-100) \log (1-25/n)$.
The terms with $\log n$ cancel out, so you get
$\log f(n) = n(4 \log 4 - 2 \log 2 - 3 \log 3) - 100 \log 4n + (4n-100) \log (1-25/n)$.
The first term grows linearly. The second term grows logarithmically. The third term approaches $-100$ as $n \to \infty$. So the whole thing grows linearly as $n \to \infty$; in particular since $4 \log 4 - 2 \log 2 - 3 \log 3 = \log (256/108) > 0$, the whole thing goes to infinity, as desired.