<s>[INST] Prove a limit using the epsilon-delta definition $\lim_{x\to-2}\frac{4x-1}{x+1} = 9$
Given $\epsilon>0$, $$(\exists \delta(\epsilon)>0) \left( |x+2|<\delta \implies \left|{\frac{4x-1}{x+1} - 9}\right| < \epsilon \right)$$
So, if $|x+2|<\delta$, then:
$$\left|{\frac{4x-1}{x+1} - 9}\right| = \left|{\frac{4x-1-9(x+1)}{x+1}}\right| = \left|{\frac{-5x-10}{x+1}}\right| = \left|{\frac{-1 (5x+10)}{x+1}}\right| = {\frac{|-1||5x+10|}{|x+1|}} = {\frac{5|x+2|}{|x+1|}}$$
I've found the expression $5|x+2|$ in the numerator:
$|x+2| < \delta = 1/2$
$-1/2<x+2<1/2$
$-5/2<x<-3/2$
$-3/2<x+1<-1/2 \implies$ ?
A colleague told me to leave both members of this inequality as positives. Why do we need both sides to be positives?
 [/INST] You're right to want to bound values of $x$ away from $-1$, and to do so by $\frac{1}{2}$ is fine.  The inequality you obtain by requiring $x$ to be within $\frac{1}{2}$ of $2$ is
$$-\frac{5}{2}<x<-\frac{3}{2}$$
which you have written.  Then, as you've written, we have 
$$-\frac{3}{2}<x+1<-\frac{1}{2}$$
Notice that the above inequality implies that $|x+1|>\frac{1}{2}.$  It follows that
$$\frac{5|x+2|}{|x+1|}<10|x+2|$$
So let $\delta(\epsilon)<\min\{\frac{1}{2},\frac{\epsilon}{10}\}$.
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