limits - 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.

Answer

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 $\delta$.