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

Question

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.

Answer 1

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 $\delta\leq\min\{\dfrac{1}{15}\varepsilon,\dfrac12\}$.

Answer 2

Hint: Recall that $0<a<b$ implies that $\frac{1}{a}>\frac{1}{b}>0$. So,

$$0<\vert x-2 \vert<\delta$$ $$\Longrightarrow \ 1<\vert x-2 \vert+1<\delta+1 $$ $$\Longrightarrow \ 1>\frac{1}{\vert x-2\vert+1}>\frac{1}{\delta+1} $$