How to Show the following converges to $e^{\frac{t^2}{2}}$ How to prove that $$\lim_{n\to\infty}\left[\left[e^{t\sqrt{\frac{1-p}{np\vphantom{()}}}}-1-t\cdot\sqrt{\frac{1-p}{np}}-\frac{1}{2}t^2\left(\frac{1-p}{np}\right)\right]\cdot p \\+\left[e^{-t\sqrt{\frac{p}{n(1-p)}}}-1+t\cdot\sqrt{\frac{p}{n{(1-p)}}}-\frac{1}{2}\cdot t^2\left(\frac{p}{n(1-p)}\right)\right](1-p)+\left(1+\frac{t^2}{2n}\right)\right]^n$$ converges to $e^{\frac{t^2}{2}}$  . where $0 \leq p\leq 1$ and $t$ is a parameter which can take any real value
 A: Reacll that: $e^{x}=1+x+\frac{x^{2}}{2}+o(x^{2})$ for small $x$ which corresponds to large $n$. Taking $x=t\sqrt{\frac{1-p}{np}}$ and $x=-t\sqrt{\frac{p}{n(1-p)}}$ we see that the first two terms are $o(x^{2})$. Notice also that as $n\to\infty$ we have $(1+\frac{\frac{t^{2}}{2}}{n})^{n}\to e^{\frac{t^{2}}{2}}$. So we have:
$((1+\frac{\frac{t^{2}}{2}}{n})+o(t^{2}(\frac{1-p}{np}))+o(t^{2}\frac{p}{n(1-p)}))^{n}=(1+\frac{\frac{t^{2}}{2}}{n})^{n}+o(t^{2}\frac{1-p}{p})+o(t^{2}\frac{p}{1-p})$
which tends to $e^{\frac{t^{2}}{2}}$.
To make the notation simpler let $u=t\sqrt{\frac{1-p}{p}}$ and $v=t\sqrt{\frac{p}{n(1-p)}}$. Then the limit we are interested in is:
$\lim_{n\to\infty}((e^{\frac{u}{\sqrt{n}}}-1-\frac{u}{\sqrt{n}}-\frac{u^{2}}{2}\frac{1}{n})p+(e^{\frac{-v}{\sqrt{n}}}-1+\frac{v}{\sqrt{n}}-\frac{v^{2}}{2n})(1-p)+(1+\frac{\frac{t^{2}}{2}}{n}))^{n}$
Taking exponentials it suffices to show the following tends to $\frac{t^{2}}{2}$:
$\lim_{n\to\infty} n\ln((e^{\frac{u}{\sqrt{n}}}-1-\frac{u}{\sqrt{n}}-\frac{u^{2}}{2}\frac{1}{n})p+(e^{\frac{-v}{\sqrt{n}}}-1+\frac{v}{\sqrt{n}}-\frac{v^{2}}{2n})(1-p)+(1+\frac{\frac{t^{2}}{2}}{n}))$
$=\lim_{n\to\infty}\bigg(\frac{\ln\bigg(\frac{(e^{\frac{u}{\sqrt{n}}}-1-\frac{u}{\sqrt{n}}-\frac{u^{2}}{2}\frac{1}{n})p}{\frac{u^{2}}{n}}\frac{u^{2}}{n}+\frac{(e^{\frac{-v}{\sqrt{n}}}-1+\frac{v}{\sqrt{n}}-\frac{v^{2}}{2n})(1-p)}{\frac{v^{2}}{n}}\frac{v^{2}}{n}+(1+\frac{\frac{t^{2}}{2}}{n})\bigg)}{(\frac{(e^{\frac{u}{\sqrt{n}}}-1-\frac{u}{\sqrt{n}}-\frac{u^{2}}{2}\frac{1}{n})p}{\frac{u^{2}}{n}}\frac{u^{2}}{n}+\frac{(e^{\frac{-v}{\sqrt{n}}}-1+\frac{v}{\sqrt{n}}-\frac{v^{2}}{2n})(1-p)}{\frac{v^{2}}{n}}\frac{v^{2}}{n}+(\frac{\frac{t^{2}}{2}}{n}))}\bigg)\cdot$
$\lim_{n\to\infty}\bigg({(\frac{(e^{\frac{u}{\sqrt{n}}}-1-\frac{u}{\sqrt{n}}-\frac{u^{2}}{2}\frac{1}{n})p}{\frac{u^{2}}{n}}u^{2}+\frac{(e^{\frac{-v}{\sqrt{n}}}-1+\frac{v}{\sqrt{n}}-\frac{v^{2}}{2n})(1-p)}{\frac{v^{2}}{n}}v^{2}+(\frac{t^{2}}{2}))}\bigg)$
Notice that as $x\to0$ then $\frac{e^{x}-1-x-\frac{x^{2}}{2}}{x^{2}}\to0$ and $\frac{\ln(1+x)}{x}\to1$ which corresponds to $n$ tending to $\infty$. Notice that the final limit is a product of two limits. I had to separate them so that it would fit properly.
