Proof of the Central Limit Theorem using moment generating functions Below is a method of proving the Central Limit Theorem using moment generating functions.  
Let $$X_{1},X_{2},...,X_{n}$$ be a sequence of i.i.d. random variables with expected value and variance $$E(X_{i}) = \mu < \infty,  Var(X_{i})=\sigma ^{2}< \infty.$$
Now let 
$$Z_{n}=\frac{\overline{X}-\mu }{\frac{\sigma }{\sqrt{n}}} = \frac{X_{1}+X_{2}+...+X_{n}-n\mu }{\sigma \sqrt{n}}.$$
We want to show that 
$$\lim_{n \to \infty} M_{Z_{n}}(t)=e^{\frac{t^{2}}{2}}$$
where $M_{X}(t)$ is the moment generating function over some finite interval.  In order to prove this, we can define a new random variable, $Y_{i}$, which is the normalized version of $X_{i}$.  Thus, 
$$Y_{i}=\frac{X_{i}-\mu }{\sigma }.$$
Then, we can say that $Y_{i}$ is i.i.d. with expected value and variance 
$$E(X_{i}) =0,  Var(X_{i})=1.$$
Using this information, we have $$Z_{n}=\frac{\overline{Y}-\mu }{\frac{\sigma }{\sqrt{n}}} = \frac{Y_{1}+Y_{2}+...+Y_{n} }{\sqrt{n}}.$$
Finding the moment generating function gives 
$$M_{Z_{n}}(t)=E[e^{t\frac{Y_{1}+Y_{2}+...+Y_{n} }{\sqrt{n}}}] =E[e^{t\frac{Y_{1}}{\sqrt{n}}}]\cdot E[e^{t\frac{Y_{2}}{\sqrt{n}}}]\cdot ...\cdot E[e^{t\frac{Y_{n}}{\sqrt{n}}}]= M_{Y_{1}}(\frac{t}{\sqrt{n}})^{n}.$$
Lastly, 
$$\lim_{n \to \infty} M_{Z_{n}}(t)=\lim_{n \to \infty} M_{Y_{1}}(\frac{t}{\sqrt{n}})^{n} = e^{\frac{t^{2}}{2}}.$$
This concludes the proof.  However, how does one show analytically that this final limit does indeed equal 
$$e^{\frac{t^{2}}{2}}?$$
 A: By Taylor's theorem,
$$M_{Y_1}(s) = E[\exp(sY_1)] = 1 + s E[Y_1] + \frac{s^2}{2} E[Y_1^2] + s^2 h(s) = 1 + \frac{s^2}{2} + s^2 h(s), \qquad \text{where $h(s) \to 0$ as $s \to 0$},$$
where the last step uses $E[Y_1]=0$ and $\text{Var}(Y_1) = 1$.
Thus
$$M_{Y_1}(t/\sqrt{n})^n = \left(1 + \frac{t^2/2}{n} + \frac{t^2}{n} h(t^2/n)\right)^n \to e^{t^2/2}.$$
[The expression in parentheses is asymptotically equivalent to $1+\frac{t^2/2}{n}$, so the last step follows by recalling $(1+\frac{x}{n})^n \to e^x$.]

Response to comment:
For $x$ near $0$, we have $\log(1+x) = x(1+g(x))$ where $g(x) \to 0$ as $x \to 0$.
\begin{align}
&\lim_{n \to \infty} n \log M_{Y_1}(t/\sqrt{n})
\\
&= \lim_{n \to \infty} n(1-M_{Y_1}(t/\sqrt{n}))[1+g(1-M_{Y_1}(t/\sqrt{n}))]
\\
&= \lim_{n \to \infty} \left\{n\left(\frac{t^2/2}{n} + \frac{t^2}{n}h(t^2/n)\right)
\left[1+g\left(\frac{t^2/2}{n} + \frac{t^2}{n}h(t^2/n)\right)\right]\right\}
\\
&= \lim_{n \to \infty} n\left(\frac{t^2/2}{n} + \frac{t^2}{n}h(t^2/n)\right)
\cdot \lim_{n \to \infty} \left[1+g\left(\frac{t^2/2}{n} + \frac{t^2}{n}h(t^2/n)\right)\right]
\\
&= \lim_{n \to \infty} n\left(\frac{t^2/2}{n} + \frac{t^2}{n}h(t^2/n)\right)
\\
&= \frac{t^2}{2} + t^2 \lim_{n \to \infty}h(t^2/n)
\\
&= \frac{t^2}{2}.
\end{align}
