Convergence in distribution to a standard normal How to show 
$$
\lim_{n\to\infty}\left(1-\frac{t^2}{2n}-\frac{t^3}{2n^{\frac32}}\right)^{-n}=e^{\Large\frac{t^2}{2}}\,\,?
$$
LHS is expanded and approximated form of a moment generating function.
RHS is MGF of a standard normal.
This is proven in WolframAlpha, but it did not give me any steps. And I dont know how to work this out. Thanks guys. 
 A: Here is an approach. Let 
$$ u = \left(1-\frac{t^2}{2n}-\frac{t^3}{2n^{\frac32}}\right), $$
then we have
$$ (1-u)^{-n}= e^{-n \ln(1-u)} = e^{-n (-u-u^2/2-\dots)}\longrightarrow_{n\to \infty}  e^{\Large\frac{t^2}{2}}.  $$
Just work out the details to see what's going on.
Note: We used the power series 

$$\ln(1-u)= -u-\frac{u^2}{2}-\dots\,. $$

A: Given the status of the question that it was once posted here and already answered. Here is another way of solving it: Consider taking $Ln$ of the expression: but first we can simplify the notation a little bit by letting: $h = \dfrac{t^2}{2}$, and $k = \dfrac{t^3}{2}$, and also $m = \dfrac{1}{n}$, and let $f(n)$ be the expression whose limit you are finding. Then:
$\displaystyle \lim_{n \to \infty} Ln\left(f(n)\right) = \displaystyle \lim_{n \to \infty}\left(-n\cdot Ln\left(1 - \dfrac{t^2}{2n} - \dfrac{t^3}{2n^{\frac{3}{2}}}\right)\right) = -\displaystyle \lim_{m \to 0} \dfrac{Ln\left(1 - hm - km^{\frac{3}{2}}\right)}{m} = -\displaystyle \lim_{m \to 0} \dfrac{-h - 1.5km^{0.5}}{1 - hm - km^{1.5}} = h$, by L'hospitale rule. So:
$\displaystyle \lim_{n \to \infty} f(n) = e^{h} = e^{\frac{t^2}{2}}$
