Sequence of normal random variables with mean 0 and variance $n$ Is it true that for any $x \in \Bbb R$, 
$$\lim_{n\to \infty} \int_{-\infty}^{x}\frac{1}{\sqrt{2\pi n}}e^{-\frac{y^2}{2n}}dy=\frac{1}{2}$$
 A: yes
you can view it has the probability that a normal of variance $n$ and mean $0$ is in the interval $(-\infty,x).$ This probability is $0.5$ plus the probability that the variable is in $(0,x)$ if $x>0.$ 
The probability that such a normal is in $(0,x)$ is the same as the probability that a standard normal is $(0,x/\sqrt{n})$ which goes to zero as $n \to \infty.$ 
Similarly if $x<0.$
A: Take $y=u\sqrt{2n}
 $. We have (assuming $x>0
 $) $$\frac{1}{\sqrt{2\pi n}}\int_{-\infty}^{x}e^{-y^{2}/2n}dy=\frac{1}{\sqrt{\pi}}\int_{-\infty}^{x/\sqrt{2n}}e^{-u^{2}}du=\frac{1}{\sqrt{\pi}}\left(\int_{-\infty}^{0}e^{-u^{2}}du+\int_{0}^{x/\sqrt{2n}}e^{-u^{2}}du\right)=\frac{1}{2}\left(1+\textrm{erf}\left(\frac{x}{\sqrt{2n}}\right)\right)
 $$ where $\textrm{erf}\left(x\right)
 $ is the error function. Hence, using the knonw value $$\textrm{erf}\left(0\right)=0
 $$ we can conclude. If $x=0
 $ we have trivially $$\int_{-\infty}^{0}e^{-u^{2}}du=\frac{\sqrt{\pi}}{2}
 $$ and if $x<0
 $ we have $$\frac{1}{\sqrt{\pi}}\int_{-\infty}^{x/\sqrt{2n}}e^{-u^{2}}du=\frac{1}{\sqrt{\pi}}\left(\frac{\sqrt{\pi}}{2}\textrm{erfc}\left(\frac{-x}{\sqrt{2n}}\right)\right)
 $$ where $\textrm{erfc}\left(x\right)
 $ is the complementary error function. Again, we have that holds $$\textrm{erfc}\left(0\right)=1
 $$ and this complete the proof.
