$X_{n}\sim \operatorname{Bin}(n,p_{n})$. Show $\frac{X_{n}-np_{n}}{\sqrt{np_{n}}}\rightarrow N(0,1)$ if $p_{n}\rightarrow 0$ and $np_{n} \rightarrow \infty$ when $n\rightarrow \infty$.
$f_{X_{n}}(k)=\frac{n!}{(n-k)!k!}p_{n}^{k}(1-p_{n})^{n-k}$
Using Stirling's formula:
$n!=n^n e^{-n}\sqrt{2\pi n}\Big[1+\mathcal{O}(\frac{1}{n})\Big]$
$f_{X_{n}}(k)=\frac{n^{n}e^{-n}\sqrt{2\pi n}\Big[1+\mathcal{O}(\frac{1}{n})\Big]}{(n-k)^{n-k}e^{-(n-k)}\sqrt{2\pi (n-k)} \Big[1+\mathcal{O}(\frac{1}{n-k})\Big](k^{k}e^{-k}\sqrt{2\pi k} \Big[1+\mathcal{O}(\frac{1}{k})\Big])}p_{n}^{k}(1-p_{n})^{n-k}$
$=\frac{n^{n}e^{-n}\sqrt{2\pi n}\Big[1+\mathcal{O}(\frac{1}{n})\Big]}{e^{-(n-k)}\sqrt{2\pi (n-k)}\Big[1+\mathcal{O}(\frac{1}{n-k})\Big](e^{-k}\sqrt{2\pi k}\Big[1+\mathcal{O}(\frac{1}{k})\Big])}(\frac{p_{n}}{k})^{k}(\frac{1-p_{n}}{n-k})^{n-k}$
$=\frac{\sqrt{n}}{\sqrt{2\pi k(n-k)}}(\frac{np_{n}}{k})^{k}(\frac{n(1-p_{n})}{n-k})^{n-k}\Big[1+\mathcal{O}(\frac{1}{n})\Big]$
$t=\frac{k-np_{n}}{\sqrt{np_{n}}}$
$k=np_{n}+\sqrt{np_{n}}t$
$n-k=n-(\sqrt{np_{n}}t+np_{n})=n(1-p_{n})-\sqrt{np_{n}}t$
$L=\lim_{n\rightarrow \infty} (\frac{np_{n}}{k})^{k}\cdot (\frac{n(1-p_{n})}{n-k})^{n-k}\\ \ln L=\lim_{n\rightarrow \infty} \ln\Big[(\frac{np_{n}}{k})^{k}\cdot (\frac{n(1-p_{n})}{n-k})^{n-k}\Big]\\ \Leftrightarrow \lim_{n\rightarrow \infty} \Big[k\ln(\frac{np_{n}}{k})+ (n-k)\ln(\frac{n(1-p_{n})}{n-k}) \Big] \\ \lim_{n\rightarrow \infty} \ln\Big[(\frac{np_{n}}{k})^{k}\cdot (\frac{n(1-p_{n})}{n-k})^{n-k}\Big]\\ \Leftrightarrow \lim_{n\rightarrow \infty} \Big[-k\ln(\frac{k}{np_{n}})- (n-k)\ln(\frac{n-k}{n(1-p_{n})}) \Big]\\ \Leftrightarrow \lim_{n\rightarrow \infty} \Big[-k\ln(\frac{k}{np_{n}})- (n-k)\ln(\frac{n-k}{n(1-p_{n})}) \Big]\\ \Leftrightarrow \lim_{n\rightarrow \infty} \Big[-(np_{n}+\sqrt{np_{n}}t)ln(\frac{(np_{n}+\sqrt{np_{n}}t)}{np_{n}})- (n(1-p_{n})-\sqrt{np_{n}}t)\ln(\frac{n(1-p_{n})-\sqrt{np_{n}}t}{n(1-p_{n})}) \Big]\\ \Leftrightarrow \lim_{n\rightarrow \infty} \Big[-(np_{n}+\sqrt{np_{n}}t)\ln(1+\frac{\sqrt{np_{n}}}{np_{n}}t)- (n(1-p_{n})-\sqrt{np_{n}}t)\ln(1-\frac{\sqrt{np_{n}}t}{n(1-p_{n})}) \Big]$
$\ln (1+x)=x+\frac{x^{2}}{2}+\mathcal{O}(x^{3})$
$\lim_{n\rightarrow \infty}\Big(-(np_{n}+\sqrt{np_{n}}t)\Big[\frac {t}{\sqrt{np_{n}}}-\frac{1}{2}\cdot \frac{t^{2}}{np_{n}}+\mathcal{O}(\frac{t^{3}}{n\sqrt{n}})\Big]\\-(n(1-p_{n})-\sqrt{np_{n}}t)\Big[-\frac{\sqrt{np_{n}}t}{n(1-p_{n})}+\frac{1}{2}\cdot \frac{np_{n}}{n^{2}(1-p_{n})^{2}}t^{2}+\mathcal{O}(\frac{t^{3}}{n\sqrt{n}})\Big]\Big)\\ =\lim_{n\rightarrow \infty}\Big(-\sqrt{np_{n}}t+\frac{1}{2}\cdot t^{2}-t^{2}+\mathcal{O}(\frac{t^{3}}{n\sqrt{n}})\\ +\sqrt{np_{n}}t+\frac{1}{2}\cdot \frac{np_{n}}{n(1-p_{n})}t^{2}-\frac{np_{n}}{n(1-p_{n})}t^{2}+\mathcal{O}(\frac{t^{3}}{n\sqrt{n}})\Big)\\ =\lim_{n\rightarrow \infty}\Big(-\frac{1}{2} \cdot t^{2}-\frac{1}{2}\cdot \frac{np_{n}}{n(1-p_{n})}t^{2}+\mathcal{O}(\frac{t^{2}}{n\sqrt{n}})\Big)\\ =\lim_{n\rightarrow \infty}\Big(-\frac{1}{2} \cdot t^{2}(1+\frac{np_{n}}{n(1-p_{n})})\Big)\\ =\lim_{n\rightarrow \infty}\Big(-\frac{1}{2} \cdot t^{2}(\frac{n(1-p_{n})}{n(1-p_{n})}+\frac{np_{n}}{n(1-p_{n})})\Big)\\ =\lim_{n\rightarrow \infty}\Big(-\frac{1}{2} \cdot t^{2}(\frac{1}{1-p_{n}})\Big)\\ =\lim_{n\rightarrow \infty}\Big(-\frac{t^{2}}{2(1-p_{n})}\Big)\\ \text{as}\qquad n\rightarrow \infty, \quad p_{n} \rightarrow 0, \quad \text{thereby}\\ \lim_{n\rightarrow \infty}\Big(-\frac{t^{2}}{2(1-p_{n})}\Big)=-\frac{t^{2}}{2}\\ \ln L=-\frac{t^{2}}{2}\Leftrightarrow L=e^{-\frac{t^{2}}{2}}\\$
This all worked out. Now to the square root of the equation:
$\frac{\sqrt{n}}{\sqrt{2\pi k(n-k)}} = \frac{\sqrt{n}}{\sqrt{2\pi (np_{n}+\sqrt{np_{n}}t)(n(1-p_{n})-\sqrt{np_{n}}t)}}\\ =\frac{\sqrt{n}}{\sqrt{2\pi(n^{2}p_{n}(1-p_{n})-np_{n}\sqrt{np_{n}}t+n\sqrt{np_{n}}(1-p_{n})t-np_{n}t^{2}}}\\ =\frac{\sqrt{n}}{\sqrt{2\pi(n^{2}p_{n}(1-p_{n})+n\sqrt{np_{n}}t(1-2p_{n})-np_{n}t^{2})}}\\ =\frac{1}{\sqrt{2\pi(np_{n}(1-p_{n}))}}\Big[1+\mathcal{O}(\frac{1}{n})\Big]$
This time the $np_{n}$ did not cancel out. Did I miss some way to cancel it out to get the square root part of the equation to $\frac{1}{\sqrt{2\pi}}$ when $p_{n}\rightarrow 0$ (and $np_{n}\rightarrow \infty$) when $n\rightarrow \infty$?