Basic $\delta-\epsilon$ proof? I am having some trouble  with this question. 
Prove this limit
$\lim_{x \to 1}  \frac{(x+2)(x-1)(x+1)}{(2x-3)} = 0$
What I did was find a Sigma of epsilon that is less than $\delta<\frac{1}{2}$ and when I started to bound each of the unwanted terms I got 
$|{x+2}|<4$
$|{x+1}|<3$
$|{2x-3}|>-3$  
Now this where I have the problem when I try to find an expression for $\delta(\epsilon)$ I get  $|x-1|<\frac{\epsilon}{-4}$. I know $\epsilon>0$ so what can I do to get rid of the negative four and get a positive bound. I know also I have to do something about $|{2x-3}|>-3$  to make it positive. Are there any theorems or tricks I can use?

You have the right idea, but your bounds just aren't good enough. $|x-1|<\frac{1}{2}$ turns out not to be the best choice, because $2\cdot\frac{3}{2}-3=0$.
So instead try $|x-1|<\frac{1}{4}$. This implies that $\frac{3}{4}<x<\frac{5}{4}$, hence
$$ -\frac{3}{2}<2x-3<-\frac{1}{2} $$
so we can say that $|2x-3|>\frac{1}{2}$ in this case. Similarly, $|x+2|<\frac{13}{4}$ and $|x+1|<\frac{9}{4}$, so puuting this all together we get
$$ \frac{|x+2||x-1||x+1|}{|2x-3|}<2\cdot\frac{13}{4}\cdot \frac{9}{4}|x-1|$$
From here, given $\varepsilon>0$, it shouldn't be too hard to pick an appropriate $\delta$.
