Vtyjkyuk

I just want to verify the preliminary work of my epsilon delta proof for this one.

$$lim_xrightarrow 2fracx-1x^2-1 = frac13 $$

$underlinetextprelimnary work$

$forall epsilon>0, exists delta>0$ $s.t.$
$|x-2|<deltaRightarrow left|fracx-1x^2-1-frac13right|<epsilon$

we know
$$left|fracx-1x^2-1-frac13right| =
left|frac(x-1)(x-2)3(x-1)(x+1)right|$$

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 $delta = min(0.5, epsilon)$

$underlinetextproof$

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 $delta = min(0.5,7.5epsilon)$