Proving $\lim_{x \to -2}$ $\frac{(x-1)}{(x+1)}$ = 3 by definition Here we go. $\lim_{x \to -2}$ $\frac{(x-1)}{(x+1)}$ = 3 
Proof: $\vert \frac{x-1}{x+1} -3 \vert$ =...= $\frac{\vert-2\vert \vert x-(-2) \vert}{(x+1)}$$\lt \frac{1}{2} \vert x -(-2) \vert$...now let $\epsilon$ $\gt$ 0 arbitrary and $\delta$ = min{1,2$\epsilon$} then we have
$\frac{\vert-2\vert \vert x-(-2) \vert}{(x+1)}$$\lt \frac{1}{2} \vert x -(-2) \vert$$\lt$$\frac{1}{2} * 2\epsilon$ = $\epsilon$ so $0<\vert x-(-2) \vert< \delta$
yes or no or I need more

If your $\delta$ is $\min\{1, 2\epsilon\}$ then it is not always true that $$\frac{2|x+2|}{|x+1|} < \frac{1}{2}|x+2|.$$
When $\epsilon$ is large, then $0 < |x + 2| < \delta = 1$ implies only that $-1 < x + 2 < 1$, or $-2 < x + 1 < 0$, so that $2 > |x+2| > 0$ and $1/2 < 1/|x+2|$ is not bounded above. The problem is $|x + 1|$ can get arbitrarily large because the bound, $\delta$, is not tight enough; for example, we can pick $x = -1.001$ and see that $0 < |x + 2| = 0.999 < 1$ is satisfied, but $2|x+2|/|x+1| = 1998$ is astronomical.
To remedy this, pick some $0 < c < 1$ and ask for $\delta = \min\{c, k\epsilon\}$ where $k$ is to be determined. 
Now if $0 < |x + 2| < \delta$, then $x + 2 < c$, or $x + 1 < c - 1 < 0$. Taking absolute values reverses the inequality: $|x+1| > |1-c| = 1-c$. Now invert: $$\frac{1}{|x+1|} < \frac{1}{1-c}.$$
Therefore, 
$$\Bigg|\frac{x-1}{x+1} - 3\Bigg| = 2\frac{|x+2|}{|x+1|} < \frac{2\delta}{1-c} \le \frac{2k\epsilon}{1-c}$$
and this is $\le \epsilon$ iff $k \le (1-c)/2$.
Everything will be neat if you pick $c = 1/2$ and $k = 1/4$.
