I'm trying to show $\lim_{x\rightarrow-1} 3x^2 -4 = 1$. It's simple but I'm quite new to these proofs and a little shaky.
I believe I have chosen a legitimate $\delta$, the logic behind which I'll lay out, but I feel like the proof itself is a little cumbersome.
Trying to bound $|(3x^2-4) -1|$, I rearrange and get
$|3(x+1)(x-1)-2| \leq 3|(x-1)(x+1)| + 2$
Assuming $\delta\leq 1$, $|x+1|<\delta \Rightarrow x<0$
So I get $3\delta(0-1) +2 = -3\delta +2< \epsilon$ and I take $\delta = \min\left(1, \frac{\epsilon-2}{3}\right)$
Now here I'm a little stuck (though, supposing I did this correctly I understand it to have been the difficult part.
So for the proof I say:
Let $\delta = \min\left(1, \frac{\epsilon-2}{3}\right)$.
Then$ |3x^2-4 - 1|<3|(x+1)(x-1)| +2 \leq 3 \frac{\epsilon-2}{3}$(Not sure what goes in here) $+2$
So I feel at this point that I'm missing something, because I have this extra term that I somehow need to eliminate to get my RHS to $\epsilon$. Am I missing something? Is my choice of $\delta$ ok? Any help would be appreciated.
There was a typo on the practice sheet which I ought to have caught. It should be
$\lim_{x\rightarrow -1} 3x^2-4 = -1.$
Then $3|(x-1)(x+1)|<\epsilon \Rightarrow |x+1|<\frac{\epsilon}{3|x-1|}$ and bounding x s.t. $x\leq 0$
$|x+1|<\frac{\epsilon}{3}$
And then the proof is straight forward with $\delta = \min(1, \frac{\epsilon}{3})$
Thanks, sorry for the dumb typo.