<s>[INST] Proof by epsilon-delta of $\lim \limits_{x \to 1} 3x^2+1=4$ I have some doubts proving the following:
$$\lim \limits_{x \to 1} 3x^2+1=4$$
My try
$(\forall \varepsilon > 0)(\exists \space \delta > 0): (0<|x-1|< \delta \implies |3x^2+1-4| < \varepsilon)$
$\implies 0<|x-1|< \delta \iff -\delta \lt x -1 \lt \delta \iff -\delta +2 \lt x+1 \lt \delta + 2 $
By transitivity:
$\implies -(\delta + 2) \lt x+1 \lt \delta + 2 \iff 0 \lt |x+1| \lt \delta + 2$
Now, working with the epsilon part:
$|3x^2+1-4| < \varepsilon \iff 3 |x-1||x+1| \lt e$
Using the fact that $|x+1| \lt \delta + 2 \iff 3|x+1| \lt 3(\delta + 2)$, and $|x-1| \lt \delta$  by transitivity:
$3|x-1||x+1| \lt 3(\delta + 2)\delta$
Then, $\varepsilon = 3(\delta + 2) \delta$ should be enough, but, if i solve the quadratic for delta:
$\delta = \frac{-6 \pm \sqrt{36 + 12\varepsilon}}{6}$, a contradiction, because in one of the solutions $\delta \lt 0 \space \forall \varepsilon \gt 0$.
After that i tried using $\delta = 1$:
$\iff |x-1| \lt 1 \implies |3x^2-3| \lt \varepsilon$
$|x-1| \lt 1 \iff  -1 \lt x-1 \lt 1 \iff 1 \lt x+1 \lt 3$
$\implies -3 \lt x+1 \lt 3 \iff |x+1| \lt 3$
With $|x-1| \lt 1$ and $|x+1| \lt 3 \iff 3|x+1| \lt 9: $
$3 |x-1||x+1| \lt \varepsilon \implies 9 |x-1| \lt \varepsilon \iff |x-1| \lt \frac{\varepsilon}{9}$.
So $\delta = \frac{\varepsilon}{9}$ should satisfy, but i used the fact that $\delta = 1.$ How can i proceed here?. I saw that i have to use $\delta = \min\{1, \frac{\varepsilon}{9}\}$, but i don't know why. Any hints?. Are my steps correct or i did something wrong?. I'm not familiarized with epsilon-delta proofs.
 [/INST] Given $\varepsilon>0$, your formula $\delta:= \min\{1, \frac{\varepsilon}{9}\}$  works! Let's verify it.
If $|x-1|<\delta$ with $\delta>0$, then
$$|3x^2+1-4|=|3(x-1)(x-1+2)|\leq 3|x-1|(|x-1|+2)<3\delta(\delta+2).$$
Now, by the above definition, $\delta\leq 1$ AND $\delta\leq \frac{\varepsilon}{9}$ which implies
$$3\delta(\delta+2)<3\cdot \frac{\varepsilon}{9}\cdot (1+2)=\varepsilon$$
and we are done.
P.S. If we take just $\delta:=\frac{\varepsilon}{9}$, then we have that $$|3x^2+1-4|<3\cdot\frac{\varepsilon}{9}\cdot (\frac{\varepsilon}{9}+2).$$
Is it true that for all $\varepsilon>0$,  $$3\cdot\frac{\varepsilon}{9}\cdot (\frac{\varepsilon}{9}+2)<\varepsilon\;?$$ 
No, because the LHS is quadratic in $\varepsilon$! Therefore we have to impose a bound on $\delta$.
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