Moment generating function of Gamma distribution I'm trying to show that as $\alpha$ tends to 0, the gamma distribution $$\Gamma(\lambda,\alpha),$$  is properly standardised, tends to the standard normal distribution. I have figured out that the moment generating function for the gamma distribution is $$\left(\frac{\lambda}{\lambda-t}\right)^\alpha.$$ Also, I've worked out that the mean and variance of a gamma random variable is $$\frac{\alpha}{\lambda}$$ and $$\frac{\alpha}{\lambda^2}$$ respectively. 
However, I am not sure how to proceed further. I tried by defining $$Z=\frac{X-\frac{\alpha}{\lambda}}{\frac{\alpha^0.5}{\lambda}}$$ and using the fact that $M_{_Z}(t) = e^{bt}M_{X}(at)$ 
However, I can't show that $$M_{_Z}(t)=e^{t^2/2}$$ which is the moment generating function of a standard normal random variable. Is this the correct way to proceed?       
 A: First of all, you seem to be using $t$ for two different purposes: a parameter 
of the Gamma distribution and the variable in the moment generating function.  These
should be completely different.  The Gamma distribution with shape parameter $k$ and rate parameter $r$ has mean $\mu = k/r$, variance $\sigma^2 = k/r^2$, and moment generating function $M_X(t) = \left(\frac{r}{r-t}\right)^k$.  The limit you should be taking is $k \to \infty$ with $r$ fixed.  The MGF of the scaled and translated variable $Y = (X-\mu)/\sigma$ is then $M_Y(t) = \left( 1 - \frac{t}{\sqrt{k}}\right)^{-k} e^{-\sqrt{k} t}$.  I suggest taking logarithms, and using a degree-2 Taylor expansion of $\ln(1 - t/\sqrt{k})$.
A: Your post starts by saying that you want to show that the Gamma distribution, when scaled properly, tends to the standard normal.  If you need to prove this by using an explicit moment generating function argument, then you should follow the procedure described by Robert Israel.  
If, however, you can use standard facts, then you will know that a Gamma is a sum of independent exponentials with the same parameter, and then all one needs to do is to quote the Central Limit Theorem. Indeed, if a moment generating function argument is to be used, the general argument is really not very different from the specific argument for the Gamma distribution.
