In order to prove that the sample $p$-percentile $x_p, p \in [0,1]$ from a sample of $n$ is asymptotically normally distributed as $n\to\infty$, it is necessary to show the following two limits. They are given in my reading without proof.
\begin{align} (*) \;\;\; \ln\left( \frac{1}{\sqrt n} \frac{n!}{(np-1)!(n-np)!} p^{np-0.5} (1-p)^{n-np+0.5} \right) \to \mathrm{a\ constant} \\ (**) \;\;\; (np-1) \ln \left( 1+\frac{z\sqrt{1-p}}{\sqrt{np}} \right) + (n-np)\ln\left( 1-\frac{z\sqrt p}{\sqrt{n(1-p)}} \right) \to -\frac{z^2}{2} \end{align}
Assume that $np$ is an integer. I could not figure out how these results are obtained. Could anyone help me, please? Thank you!
$$\frac{1}{\sqrt{2\pi}}\times \frac{1}{e}\times\frac{n^n}{(np-1)^{np-0.5}(n(1-p))^{n(1-p)+0.5}}\times p^{np-0.5}(1-p)^{n(1-p)+0.5}= \frac{1}{\sqrt{2\pi}}\times \frac{1}{e}\times\Big(\frac{np}{np-1}\Big)^{np-0.5} \rightarrow \frac{1}{\sqrt{2\pi}}\times \frac{1}{e}\times e$$
$$(np-1)\Bigg(\frac{z}{\sqrt{n}}\sqrt{\frac{1-p}{p})}-\frac{z^2}{2n}\frac{1-p}{p} +O(n^{-3}) \Bigg)+n(1-p)\Bigg(-\frac{z}{\sqrt{n}}\sqrt{\frac{p}{1-p}}-\frac{z^2}{2n}\frac{p}{1-p}+O(n^{-3})\Bigg)=\Bigg(-\frac{z^2}{2}(1-p)-\frac{z^2}{2}p\Bigg)+o(1)\rightarrow -\frac{z^2}{2}$$