# Second half of epsilon-delta limit proof

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.

• In real life, $$\lim_{x\rightarrow-1} 3x^2 -4 = \color{red}{-1}$$ – Did Nov 14 '16 at 14:06
• @Did perhaps he took the red pill. Who's to say this is the real world? – Brevan Ellefsen Nov 14 '16 at 14:09
• Not quite. Your $|f(x) - L|$ should be $|3x^2 - 3|$, not $|3x^2 - 5|$. – Nitin Nov 14 '16 at 14:09
• It was literally written incorrectly on the practice sheet I was given and I didn't think to check. sigh. Is that why my proof doesn't work? – BenL Nov 14 '16 at 14:10
• That's frustrating... but yeah, you can't prove something that isn't true, after all. This should be easier though because you have 2 easy linear terms, i.e it becomes $3|x+1||x-1| < 3 \delta (2 + \delta)$. – Nitin Nov 14 '16 at 14:14

## 1 Answer

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.