1. Limits

Use the epsilon-delta definition of limit to prove $\lim_{x\to -1}\frac{x^2-3}{x+2}=-2$

2. $\left|\frac{x^2- 3}{x+ 2}- (-2)\right|= \left|\frac{x^2+ 2x+ 1}{x+2}\right|$ $=\left|\frac{(x+ 1)^2}{x+2}\right|$.

Does that help?

3. not really, thats where i got stuck. i dunno if this is rite but...

$|x+1|<\frac{1}{2}\Rightarrow \frac{1}{2}
$\frac{|x+1|^2}{|x+2|}<2|x+1|^2$

and then set delta=sqrt(epsilon/2)