Show convergence in distribution by the continuity theorem So the problem I'm about to solve is to show that:
$X \in \Gamma(a,b)$. Show that
\begin{equation}
\frac{X-E[X]}{\sqrt{Var(X)}} \xrightarrow{d} N(0,1)
\end{equation}
as $a \rightarrow \infty$, by using the continuity theorem.
I've made it this far:
\begin{equation}
\left(\frac{\exp{\left(-\frac{it}{\sqrt{a}}\right)}}{1-\frac{it}{\sqrt{a}}}\right)^{a}
\end{equation}
and now I want to show the limit when $a \rightarrow \infty$.
I have checked with a WolframAlpha, and the limit goes to $e^{-\frac{t^{2}}{2}}$, as expected, but I have no clue how to show this.
 A: Recall the key result that if $z(a)\to w$ in $\mathbb C$ then $$\left(1+\frac{z(a)}a\right)^a\to\mathrm e^w,$$ hence the idea is to check that $$\frac{\exp{\left(-\frac{it}{\sqrt{a}}\right)}}{1-\frac{it}{\sqrt{a}}}=1+\frac{z(a)}a,\tag{$\ast$}$$ for some $z(a)$ converging to some finite limit $w$ when $a\to\infty$. To do so, asymptotic expansions up to order $1/a$ suffice. Introducing $$s(a)=-\frac{it}{\sqrt{a}},$$ one gets $s(a)\to0$. Furthermore, when $z\to0$ in $\mathbb C$, $$\mathrm e^z=1+z+\frac{z^2}2+o(z^2),$$ hence the LHS of $(\ast)$ is $$\frac{\mathrm e^{-s(a)}}{1-s(a)}=\frac{1-s(a)+\frac{s(a)^2}2+o(s(a)^2)}{1-s(a)}=1+\frac{\frac{s(a)^2}2+o(s(a)^2)}{1-s(a)}=1+\frac{s(a)^2}2+o(s(a)^2),$$ thus, $(\ast)$ holds with $$z(a)=a\left(\frac{s(a)^2}2+o(s(a)^2)\right)=-\frac{t^2}2+o(1).$$ Finally, $z(a)\to w$ with $w=-\frac12t^2$, hence the proof is complete.
A: Hint: You could consider  
$$\exp(-x) = 1 - x + \dfrac{x^2}{2!} - \dfrac{x^3}{3!} +\cdots$$  
and 
$$(1+x)^k = 1 + kx + \dfrac{k(k-1)}{2!}x^2+ \cdots$$  
and 
$$\lim_{y \to \infty} \left(1-\frac{x}{y}\right)^y =\exp(-x)$$  
