I just want to verify the preliminary work of my epsilon delta proof for this one.
$$\lim_{x\rightarrow 2}{\frac{x-1}{x^2-1}} = \frac{1}{3} $$
$\underline{\text{prelimnary work}}$
$\forall \epsilon>0, \exists \delta>0$ $s.t.$ $|x-2|<\delta\Rightarrow \left|\frac{x-1}{x^2-1}-\frac{1}{3}\right|<\epsilon$
we know $$\left|\frac{x-1}{x^2-1}-\frac{1}{3}\right| = \left|\frac{(x-1)(x-2)}{3(x-1)(x+1)}\right|$$
now we take $x>1$ to cancel $x-1$; we pick $\delta<a$ such that $2-a>1$ i.e. $0<a<1$; therefore $\delta< 0.5$ will do. Thus, $\delta<0.5$ therefore $|x-2|<\delta<0.5$ $$1.5<x<2.5$$
now we have $$\left|\frac{(x-1)(x-2)}{3(x-1)(x+1)}\right| = \left|\frac{(x-2)}{3(x+1)}\right|<|x-2|<\epsilon $$ hence $\delta = \min(0.5, \epsilon)$
$\underline{\text{proof}}$
take $\delta = 0.5$ or $\delta = \epsilon ...$
is there anything wrong with my proof? Crtisim is highly appreciated
Your proof is very good. You have explained every step very clearly.
Note that at $$\left|\frac{(x-2)}{3(x+1)}\right|<|x-2|<\epsilon$$
You could have done a little bit better by considering $3(x+1)>7.5$
Thus we could have said $$\left|\frac{(x-2)}{3(x+1)}\right|<|x-2|/{7.5}<\epsilon$$ Which makes your $\delta = \min(0.5,7.5\epsilon)$