Pointwise convergence of the cdf of normal random variables For a sequence $X_1, X_2, \dots $, Let $F_n(x)$ denote the cdf of $X_n$. 
Suppose our sequence is $X_n \sim N(0,n) $ then for all $x$ the point-wise limit of $F_n(x)$ is $\frac{1}{2}$.
How would one prove this?
 A: More generally, let $X$ be any random variable with distribution $F$ having unit variance.  Letting $(\mu_n)$ and $(\sigma_n)$ be any sequences of numbers, set
$$X_n = \sigma_n X + \mu_n.$$
Define $F_n$ to be the distribution of $X_n$.  Suppose that as $n\to\infty$, $$\sigma_n\to\infty$$ and $$\mu_n / \sigma_n\to z.$$  If $F$ is continuous in a neighborhood of $-z$, then the pointwise limit of $F_n(x)$ must be $F(-z)$.
Intuitively, this is because


*

*The distributions of the $X_n$ are becoming more and more spread out but

*Relative to the spreads $\sigma_n$, any fixed number $x$ is becoming ever closer to $-z$.
Thus the pointwise limit ought to be $F(-z)$.
It remains to make this intuition rigorous.  From the basic definitions and a little bit of algebra, observe that 
$$F_n(x) = \Pr(X_n \le x) = \Pr(\sigma_n X + \mu_n \le x) = \Pr\left(X \le \frac{x-\mu_n}{\sigma_n}\right) = F\left(\frac{x-\mu_n}{\sigma_n}\right).$$
The assumptions about the limiting values of $\mu_n/\sigma_n$ and $\sigma_n$ were made specifically to imply  $\lim_{n\to \infty} (x-\mu_n)/\sigma_n = -z$.  Applying the assumed continuity of $F$ in a neighborhood of $-z$ finishes the demonstration.  (The details, which are straightforward, are left to the reader because this is a self-study question that asks for guidance and intuition rather than a complete answer.)
Application of this result to the standard Normal distribution $F$, $(\mu_n) = (0)$, and $\sigma_n=(n)$ answers the question as stated, because $F(-0) = 1/2$.
