<s>[INST] Computing $\text{lim}_{n \rightarrow \infty}\frac{n!}{n^{\lfloor y\sqrt{n} \rfloor}(n-\lfloor y \sqrt{n}\rfloor)!}$ We fix $y \ge 0$. What is the following limit?
$$\text{lim}_{n \rightarrow \infty}\frac{n!}{n^{\lfloor y\sqrt{n} \rfloor}(n-\lfloor y \sqrt{n}\rfloor)!}$$
My attempt so far is as follows:
By Stirling's formula, $n! \approx e^{-n}n^{n+1/2} \sqrt{2\pi}$, so that we can compute:
$$\text{lim}_{n \rightarrow \infty}\frac{e^{-n}n^{n+1/2} \sqrt{2\pi}}{n^{\lfloor y\sqrt{n} \rfloor}e^{-n+\lfloor y\sqrt{n} \rfloor}(n-\lfloor y\sqrt{n} \rfloor)^{n-\lfloor y\sqrt{n} \rfloor+1/2} \sqrt{2\pi}} \\ 
 = \text{lim}_{n \rightarrow \infty}e^{-\lfloor y\sqrt{n} \rfloor} \left( \frac{n}{n-\lfloor y\sqrt{n} \rfloor} \right)^{n-\lfloor y\sqrt{n} \rfloor+1/2}$$
upon cancellation and rearrangement of terms.  This is endlessly confusing for me to compute at this point.  Does anyone know how to deal with the limit from here?  I'm asking this question as part of another: Complicated distribution theory question for reference.
 [/INST] I've chosen to ignore the floor function; you can probably modify this argument to include the floor function if you wish.
The logarithm is
\begin{align}
&-y\sqrt{n} + (n - y\sqrt{n} + 1/2) \log(\frac{n}{n-y\sqrt{n}})
\\
&=-y\sqrt{n} + (n - y\sqrt{n} + 1/2) \log(1 + \frac{y}{\sqrt{n}-y})
\\
&= -y\sqrt{n} +(n - y\sqrt{n} + 1/2) \left(\frac{y}{\sqrt{n}-y} - \frac{1}{2} \frac{y^2}{(\sqrt{n}-y)^2} + O(n^{-3/2})\right)
\\
&= -y\sqrt{n} + y(\sqrt{n} + \frac{y}{2(\sqrt{n}-y)})
 - \frac{y^2}{2} (\frac{\sqrt{n}}{\sqrt{n}-y} + \frac{1}{2(\sqrt{n} - y)^2})
 + O(n^{-1/2})
\\
&\to - y^2/2.
\end{align}
So your limit is $e^{-y^2/2}$
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