I am trying to rigorously (without a calculator) show $$\lim_{n\to \infty }\sqrt{n}\color{blue}{{{n}\choose {\Big[ np + \sqrt{np(1-p)}\,\Big]}}p^{\Big[np + \sqrt{np(1-p)} \,\Big]}{(1-p)^{\bigg(n-\Big[np + \sqrt{np(1-p)}\,\Big]\bigg)}}}=0$$ where $0<p<1$ and $[\cdot ]$ denotes the nearest integer. This limit was formed while playing with the $\color{blue}{\text{binomial mass function}}$ and while fiddling with the proof of the DeMoivre/Laplace Central Limit Theorem:
https://en.wikipedia.org/wiki/De_Moivre%E2%80%93Laplace_theorem
I am not completely convinced that the limit is 0, but after seeing some computations on Mathematica
I see that it tends to $0$ for several rational values of $p$. I want to see whether the limit is $0$ for all $p$ in $(0,1)$.
What I've tried: (1) Using ${{a}\choose{b}}\le (\frac{ae}{b})^b$ and dropping $[ \cdot]$ for simplicity (and hopefully at no cost) to get $$\sqrt{n}\Bigg(\frac{ne}{np + \sqrt{np(1-p)}}\Bigg)^{np + \sqrt{np(1-p)}}p^{np + \sqrt{np(1-p)}}(1-p)^{n-np + \sqrt{np(1-p)}} \\\le \sqrt{n}\Bigg(\frac{nep}{np(1-p)+n\sqrt{\frac{p(1-p)}{n}}(1-p) }\Bigg)^{np+ \sqrt{np(1-p)}}(1-p)^n\\\le \sqrt{n}\bigg(\frac{e}{(1-p)}\bigg)^{2np}(1-p)^n$$
but according to Mathematica the latest expression doesn't tend to 0, so I may need a better upper bound.