Note that your function can be written as
[tex]\begin{cases}\frac{2-(-x)}{2+x}&\text{if } x<0\\\frac{2-x}{2+x}&\text{if } x>0\end{cases} = \begin{cases}\frac{2+x}{2+x}&\text{if } x<0\\\frac{2-x}{2+x}&\text{if } x>0\end{cases} = \begin{cases}1&\text{if } x<0\\\frac{2-x}{2+x}&\text{if } x>0\end{cases}[/tex]
So, if x is negative, your function is identically equal to 1. In particular, the limit as x approaches -2 will be 1 as well.
