Suppose X$_n$ folowes $N(\mu,\frac{1}{n^2})$. Thus
$$ F_{X_n} = \frac {n}{\sqrt{2\pi}}\int_{-\infty}^xexp(-\frac{1}{2}n^2y^2)dy= \frac {n}{\sqrt{2\pi}}\int_{-\infty}^{nx} exp(-\frac{1}{2}z^2)dz$$.
Why does the integral boundry change?
It follow that (sorry for the format, couldnt figure out how to typeset a piecwise function here):
$$ lim_{n->\infty} F_{X_N(x)= 0 \text{ if x<0}; \frac12 \text{ if x=0}};1 \text{ if x>0}$$
Note that $lim_{n->\infty} is not a CDF. However, it does hold that
$$ F_n \text{converges in distribution to } F $$
$$ F(x) = 0 \text{ if x , 0 }; 1 \text{ if x $\geq$ 0}$$
What is the reasoning behind these conclusions ?
This exmpale is from a chapter on convergence and includes Chebychev`s Inequality, Theorem of Levy-Cramer and Lemma of Slutsky.
There is a mistake in the second expression for $F_{X_{n}}$. There is no $n$ in the last expression for it. The correct formula is $F_{X_{n}}=\frac 1 {\sqrt {2\pi}} \int_{-\infty}^{nx}exp(-\frac 1 2 z^{2})\, dz$. You get this by making the substitution $z=ny$. (As $y$ ranges from $-\infty$ to $x$, $z$ ranges from $-\infty$ to $nx$). When you take the limit of $F_{X_{n}}(x)$ you get integral over the whole line $\mathbb R$ if $x>0$, integral over the empty set if $x<0$ and integral over $(-\infty, 0)$ if $x=0$. This gives the limit as $1$,$0$ and $\frac 1 2$ respectively. This limiting function is not a distribution function because it is not continuous from the right at $0$. However, if you redefine the value at $0$ to be $1$ you will get a distribution function $F$ and $F_{X_{n}} \to F$ at all points where $F$ is continuous (namely, points other than $0$); Hence $F_{X_{n}}$ converges in distribution.