Prove the limit $\lim\limits_{x\to -2} (3x^2+4x-2)=2 $ using the $\epsilon,\delta$ definition.
Precalculations
My goal is to show that for all $\epsilon >0$, there exist a $\delta > 0$, such that $0<|x+2|<\delta\ \ \text{implies}\ |3x^2+4x-2-2|<\epsilon$
$|3x^2+4x-2-2|=|3(x+2)^2-8x-16|$
$=|3(x+2)^2-4(x+2)|$
$\leq3|x+2|^2+4|x+2|$ by triangle inequality
$<3\delta^2+4\delta$
Hence, it is sufficient to show that $3\delta^2+4\delta=\epsilon$
Proof
For all $\epsilon>0$, choose $\delta=\min\left(\sqrt{\dfrac{\epsilon}{6}},\dfrac{\epsilon}{8}\right)$
$\begin{align*}0<|x+2|<\delta\ \ \to\ \ &|3x^2+4x-2-2|<3\delta^2+4\delta\\&<3\left(\sqrt{\frac{\epsilon}{6}}\right)^2+4\delta\\&=\frac{\epsilon}{2}+4\delta\\&<\frac{\epsilon}{2}+4\frac{\epsilon}{8}\\&=\frac{\epsilon}{2}+\frac{\epsilon}{2}\\&=\epsilon\end{align*}$
Therefore proven? Hehe. Not sure this will work or not. My doubts lies in the steps.
Hence, it is sufficient to show that $3\delta^2+4\delta=\epsilon$
choose $\delta=\min\left(\sqrt{\dfrac{\epsilon}{6}},\dfrac{\epsilon}{8}\right)$
And hey, I am looking out for other possible ways to do this question too.