$\epsilon$-$\delta$ limit proof, $\lim_{x \to 2} \frac{x^{2}-2x+9}{x+1}$ Prove that $\lim\limits_{x \to 2} \frac{x^{2}-2x+9}{x+1}$ using an epsilon delta proof.
So I have most of the work done.  I choose $\delta = min{\frac{1}{2}, y}$,
$f(x)$ factors out to $\frac{|x-3||x-2|}{|x+1|}$  
But $|x-3| \lt \frac{3}{2}$ for $\delta = \frac{1}{2}$ and also $|x+1| > 5/2$ (I'll spare you the details).  
I'm not sure how to choose my y here. If I take $\lim\limits_{x \to 2} \frac{x^{2}-2x+9}{x+1}$ < $(3/5) \delta$   How do I choose my epsilon here (replace y with this) to satisfy this properly?
Thanks

I'm going to go out on a limb and guess that you're trying to show the limit is 3 and that $f(x) = {x^2 - 2x + 9 \over x + 1} - 3$. I suggest trying to translate what you've done into the fact that  $|{x^2 - 2x + 9 \over x + 1} - 3| < {3 \over 5}|x - 2|$ whenever $|x - 2| < {1 \over 2}$. 
This means that if you choose any $\epsilon < {1 \over 2}$, then you have that $|{x^2 - 2x + 9 \over x + 1} - 3| < {3 \over 5}\epsilon$ whenever $|x - 2| < \epsilon$. So, given $\epsilon$, the natural choice for $\delta$ is ${3 \over 5}\epsilon$. (you got the $\delta$ and $\epsilon$ reversed.)
Now verify that the "for every $\epsilon$ there is a $\delta$" definition is satisfied in this way. 
