Show that $\frac{1}{x+2}$ is continuous for $(-2,0]$. I must use an $ε-δ$ proof to prove the function $\frac{1}{x+2}$ is continuous for $(-2,0]$.
I set $x \in (-2,0]$, and $y\in (-2,0]$ s.t $|x-y|<δ$.
I am having a hard time determining what precisely I should set $y$ to, as 
$|\frac{1}{x+2}-\frac{1}{y+2}|=|\frac{(y+2)-(x+2)}{(x+2)(y+2)}|=\frac{|y-x|}{(x+2)(y+2)}$
and at this point I am stuck on how to proceed.

Hint
For $\vert y-x \vert \le \frac{x +2}{2}$ you have
$$-\frac{x +2}{2} \le y-x \le \frac{x +2}{2}, \text{ hence } 2 + y \ge \frac{x}{2} + 1 = \frac{x+2}{2}$$
and therefore
$$0 \le \frac{1}{2+y} \le \frac{2}{2+x}$$
Finally
$$\left\vert\frac{1}{x+2}-\frac{1}{y+2}\right\vert=\left\vert\frac{(y+2)-(x+2)}{(x+2)(y+2)}\right\vert=\frac{|y-x|}{(x+2)(y+2)} \le \frac{2}{(x+2)^2} \vert y-x\vert$$
For $\epsilon >0$ given, $\delta = \min\{\frac{x+2}{2}, \frac{\epsilon (x+2)^2}{2}\}$ should do the job.
