Prove by the limit definition that $\lim_{x\rightarrow 2}\frac{x^2-1}{x-3}=-3$ I was reviewing $\mathbb{R}-$analisys with a friend and I'm thinking about one of the questions...

Prove by the $\epsilon-\delta$ limit definition that $\lim_{x\rightarrow 2}\frac{x^2-1}{x-3}=-3$.

My answer was very long, someone could do a better answer? Does is it OK in this way? Thanks very much.
My attempt
We have $|f(x)-(-3)|=|f(x)+3|=|\frac{x^2-1}{x-3}+3|=|\frac{x^2-1+3x-9}{x-3}|=|\frac{x^2+3x-10}{x-3}|=|\frac{(x-2)(x+5)}{x-3}|  \qquad (i)$
and
$|x+5|=|x+7-2|<|x-2|+|7|=|x-2|+7  \qquad (ii)$
$ |x-3|=|x-2-1|>|x-2|+|-1|=|x-2|+1\Longrightarrow \dfrac{1}{|x-3|}<\dfrac{1}{|x-2|+1}  \qquad(iii)$
So, by (ii) and (iii)
$|\frac{(x-2)(x+5)}{x-3}|=\dfrac{|x-2||x+5|}{|x-3|}<\dfrac{|x-2|(|x-2|+7)}{|x-2|+1}  \qquad (iv)$
But also, if $|x-2|<\delta  \qquad (v)$, we have
$|\frac{(x-2)(x+5)}{x-3}|<^{(iv)}\dfrac{|x-2|(|x-2|+7)}{|x-2|+1}<^{(v)}\dfrac{\delta(\delta+7)}{\delta+1}  \qquad(vi)$
And more than that
$\dfrac{\delta(\delta+7)}{\delta+1}<\dfrac{\delta(\delta+1)}{\delta+1}=\delta \qquad (vii)$
So, if $\delta<\epsilon$, we have
 $|f(x)-(-3)|=^{(i)}|\frac{(x-2)(x+5)}{x-3}|<^{(vi)}\dfrac{\delta(\delta+7)}{\delta+1}<^{(vii)}\delta<\epsilon$, Q.E.D.

I think (i) part is correct, and assume we choose $x$ from $|x-2|<\dfrac12$ then
$$\dfrac32<x<\dfrac52$$
$$-\dfrac32<x-3<-\dfrac12$$
$$\dfrac{13}{2}<x+5<\dfrac{15}{2}$$
these show 
$$|\frac{(x-2)(x+5)}{x-3}|<\dfrac{15}{2}|x-2|.2<15\delta$$
so it is sufficient to have $\delta\leq\min\{\dfrac{1}{15}\varepsilon,\dfrac12\}$.
