How to define the dirac delta as the limit of an arbitrary probability distribution? Is it true that, for an arbitrary probability density, $\rho(x)$, that,
  $$\lim_{\alpha\to0^{+}}\frac{1}{\alpha}\rho(x/\alpha)=\delta(x),$$
where $\delta(x)$ is the dirac delta?
If this is not true please can you provide some additional minimal conditions under which it becomes so. 
(for clarity by probability density I mean a function which satisfies, $\rho(x)\ge0$ $\forall x$ and $\int_{-\infty}^{\infty}\rho(x)\mathrm{d}x=1$) 

By definition, $\frac{1}{\alpha} \rho(x/\alpha) \to \delta$ as $\alpha \to 0^+$ if $\int \frac{1}{\alpha} \rho(x/\alpha) \, \varphi(x) \, dx \to \varphi(0)$ for every $\varphi \in C_c^\infty.$
Therefore, let $\varphi \in C_c^\infty.$ Then,
$$
\int \frac{1}{\alpha} \rho(x/\alpha) \, \varphi(x) \, dx
= \{ \text{ substitution $x = \alpha y$ } \}
= \int \rho(y) \, \varphi(\alpha y) \, dy\
.
$$
Now, since $|\rho(y)\,\varphi(\alpha y)| \leq |\rho(y)|\,\sup|\varphi| \in L^1$ we can apply the dominated convergence theorem and get the limit
$$
\int \rho(y) \, \varphi(0) \, dy
= \int \rho(y) \, dy \ \varphi(0)
= \varphi(0)
.
$$
Thus, $\frac{1}{\alpha} \rho(x/\alpha) \to \delta.$
