distribution convergence N(0,σ^2/n) Consider the sequence $X_1, X_2 \dots$ where each $X_n$ has a normal distribution $\mathcal{N}(0, σ ^ 2 / n)$. Let us show that $X_n$ converges in distribution to 0. 
$$F_{X_n} (x) = \frac{1}{\sqrt{2\pi\sigma^2/n}} \int_{-\infty}^x e^{\frac{-u^2}{2(\sigma^2/n)}}du$$
My question is: Why is the limit of the distribution function as follows? how do you get there from the equation above?
$$\lim_{n\rightarrow \infty} F_{X_n} (x) =  \begin{cases} 
      0 & \text{si}\ x < 0, \\
      \frac{1}{2} & \text{si}\ x = 0, \\
      1 & \text{si}\ x > 0.
   \end{cases}
$$

Hint: with $v = u\sqrt{n}/\sigma$,
$$\frac{1}{\sqrt{2\pi \sigma^2/n}} \int_{-\infty}^x e^{-u^2 / (2 \sigma^2/n)} \, du = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{x\sqrt{n}/\sigma} e^{-v^2/2} \, dv.$$
Then take $n \to \infty$.

Instead of working with integrals, it may be easier to note that $X_n = \frac{\sigma}{\sqrt{n}} Z$ where $Z \sim N(0,1)$. So $P(X_n \le x) = P(Z \le \frac{x\sqrt{n}}{\sigma}) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{x\sqrt{n}/\sigma} e^{-v^2/2} \, dv$.
