proof distribution Function

Question

I have to show that $F=1- \exp(-x)/2$ a distribution Function on $[0, \infty)$. I become example solution, where $\lim_{x\to-\infty} F =0$. Can anybody explain why? So i see $\lim_{x\to-\infty} F = 1-\lim_{x\to-\infty} \exp(-x) /2 =1-\infty$ and $F(0)=1/2$. Thank you.

Answer 1

When it is written "a distribution function on $[0, \infty)$" it is supposed that $F(x)=0$ on the other part of line. So you have a function $$ F(x)=\begin{cases}1-\frac12e^{-x}, & x\geq 0\cr 0, & x < 0. \end{cases} $$ And $\lim\limits_{x\to-\infty} F(x)=\lim\limits_{x\to-\infty} 0 =0 $.