Dirac Delta function and normal distribution I understand the Dirac Delta is the limit of a normal distribution when the variance of the normal distribution tends to 0:
$$
\delta(x) = \lim_{v\to 0}\frac{e^{-x^2/2v}}{\sqrt{2\pi v}}
$$
Then what is the limit
$$
\lim_{v\to 0} \frac{e^{-x^2/2v}}{\sqrt{2\pi v^n}}
$$
for some integer $n$? Can we expand the normal density function, or an integral with the normal integrator, into perhaps a power series of the variance v, which is assumed to be small?  
By the way, I typed the first limit into Mathematica9, but didn't really get the Dirac Delta.
 A: The statement $$\delta(x) = \lim_{v\to 0}\frac{e^{-x^2/2v}}{\sqrt{2\pi v}}$$ means that for every $C^\infty$ smooth function $\varphi$ with compact support we have 
$$\varphi(0) = \lim_{v\to 0} \int_{-\infty}^\infty \frac{e^{-x^2/2v}}{\sqrt{2\pi v}} \varphi(x)\,dx \tag{1}$$
(I would not expect computer algebra systems to correctly handle various modes of function convergence, especially convergence in the sense of distributions.) 
Let's replace $v$ by $v^n$ under the square root in (1); I will consider real $n$, not just integers. The resulting limit can be written as
$$ \lim_{v\to 0} v^{(1-n)/2} \int_{-\infty}^\infty \frac{e^{-x^2/2v}}{\sqrt{2\pi v}} \varphi(x)\,dx \tag{2}$$
where the integral converges to $\varphi(0)$, as previously professed. There are two cases:


*

*$n<1$. Then the limit is zero. So, the functions $\lim_{v\to 0} \frac{e^{-x^2/2v}}{\sqrt{2\pi v^n}}$ converge to zero in the sense of distributions.

*$n>1$. The limit does not exist, which means $\lim_{v\to 0} \frac{e^{-x^2/2v}}{\sqrt{2\pi v^n}}$ does not exist, even in the sense of distributions.

