Proving that $\lim_{ x\to 2} (x^2 - x -3) = -1$ using epsilon and delta

Question

Can someone check if my epsilon delta proof is correct? Thanks a lot!

Answer 1

Some of your $\delta$s look just like $s$. Your professor ought to require proofs to be in $\LaTeX$; at any rate, you'll likely get a faster response here if you learn a bit of it. Mathematically, the proof is correct, but it would be more readable if you remove the "scratch work" at the beginning, including the relevant parts of it later. The preliminary computations help you come up with the proof, but are not necessarily a part of the proof itself.

Also, it would be more efficient to use the triangle inequality to estimate $|x+1|$. Here's how my proof would begin:

Given $\epsilon>0$, let $\delta=\min\{1,\epsilon/4\}$. Consider $x$ such that
$0<|x-2|<\delta$. Since $\delta\le 1$, it follows that $|x-2|\le 1$. By the triangle inequality, $$|x+1|=|(x-2)+3|\le |x-2|+3 \le 1+3=4$$ ....