Proving limit of rational function using epsilon delta definition of a limit. $\lim_{x\rightarrow 1}\dfrac{(x-1)(x+3)}{(x-2)}  =0$
I know how to deal with the nummerator, but I am having trouble bounding the denominator in a useful way. Any hints?
$*$ What I mean is that I would like to bound $x$ in such a way that I could make this claim 
$\dfrac{|x-1||x+3|}{|x-2|}< |x-1||x+3|$ because then I know how to go about it from here. 
 A: Given any $\epsilon > 0$, let $\delta = \min\{0.5, \epsilon/A\}$, where $A$ is some fixed constant that we will figure out later (I am using $0.5$ instead of $1$ because $f(x)$ has a vertical asymptote at $x = 2$). Suppose further that we know that $|x + 3| < B$ and $|x - 2| > C$. Then if $|x - 1| < \delta$, it follows that:
$$
\left| \frac{(x - 1)(x + 3)}{(x - 2)} \right|
= |x - 1| \frac{|x + 3|}{|x - 2|}
< |x - 1| \frac{B}{C}
$$
Hence, if we take $A = B/C$, then we obtain:
$$
|x - 1| \frac{B}{C} < \frac{\epsilon}{A}(A) = \epsilon
$$
as desired.

It remains to figure out what $B$ and $C$ should be. Notice that:
\begin{align*}
|x - 1| < \delta < 0.5
&\implies -0.5 < x - 1 < 0.5 \\
&\implies -1.5 < x - 2 < -0.5 \\
&\implies 0.5 < |x - 2| < 1.5
\end{align*}
Hence, we may take $C = 0.5$. Can you see what $B$ should be, and thus what $A$ is?
A: We must prove there exists a $\delta>0$ such that $|x-1|<\delta$ implies $|f(x)|<\epsilon$ for $\epsilon>0$.
$$|x-1|<\epsilon\frac{|x-2|}{|x+3|}$$
$|x-1|<1$ which means $0<x<2$ which means
$\delta=\min (\frac{2}{5}\epsilon,0)$
Therefore the limit is $0$.
