This suggestion has a lot in common with what freckles wrote.
First let \(t=x-1\), so that as \(x\to1\) we have \(t\to0\). This doesn't change the problem (much); we still want to prove the same limit.
\[\lim_{x\to1}\frac{(x-1)(x+3)}{x-2}=\lim_{t\to0}\frac{t(t+4)}{t-1}=0\]
According to the definition of the limit, we want to show that, given any \(\epsilon>0\), we can determine an appropriate \(\delta\) so that if \(|t-0|=|t|<\delta\), then we must have \(\left|\frac{t(t+4)}{t-1}-0\right|=\left|\frac{t(t+4)}{t-1}\right|<\epsilon\).
Let's try to work backwards. The strategy behind \(\epsilon-\delta\) proofs is to extract some inequality of the form \(|t|<(\text{some expression containing }\epsilon)\) from the desired \(|f(t)-L|<\epsilon\). One thing we could do is to compare \(\frac{t(t+4)}{t-1}\) to another rational expression such that the denominator can be canceled - namely, compare it to \(\frac{t(t-1)}{t-1}=t\).
From the following plot, you can ascertain (at least for some interval containing \(t=0\)) that
\[(*)\quad\quad|t|=\left|\frac{t(t-1)}{t-1}\right|\le\left|\frac{t(t+4)}{t-1}\right|\]
http://www.wolframalpha.com/input/?i=plot+Abs%5Bt%28t%2B4%29%5D%2C+Abs%5Bt%28t-1%29%5D+over+-2%2C2
You have to be careful about this statement, though, as it's not true if \(t<-\frac{3}{2}\). To avoid this problem, we agree to fix \(|t|<1\), i.e. \(-1