I'm new to $\delta, \epsilon$ proofs and not sure if I've got the hand of them quite yet.
$ \lim_{x\to -2} (2x^2+5x+3)= 1 $
$|2x^2 + 5x + 3 - 1| < \epsilon$
$|(2x + 1)(x + 2)| < \epsilon$
$|(2x + 4 - 3)(x + 2)| < \epsilon$
$|(2(x+2)^2 -3(x + 2)| \leq 2|x+2|^2 +3|x + 2| < \epsilon$ (via the triangle inequality)
Let $|x+2| < 1$
Then $\delta=\min\left(\dfrac{\epsilon}{5}, 1\right)$
and
$\lim_{x\to -2} (3x^2+4x-2)= 2$
$|3x^2 + 4x - 2 - 2| < \epsilon$
$|3x^2 - 12 + 4x + 8| < \epsilon$
$|3(x+2)(x-2) + 4(x + 2)| < \epsilon$
$|3(x+2)(x + 2 -4) + 4(x + 2)| < \epsilon$
$|3[(x+2)^2 -4(x+2)] + 4(x + 2)| < \epsilon$
$|3(x+2)^2 - 8(x+2)| \leq 3|x+2|^2 + 8|x+2| < \epsilon$
Let $|x + 2| < 1$
Then $\delta =\min\left(\dfrac{\epsilon}{11}, 1\right)$
and to make sure I'm understanding this properly, when we assert that $|x+2| < 1$, this means that $\delta \leq 1$ as well, because if $\delta \geq 1$, this would allow for $|x+2| \geq 1$, which violates that condition we just imposed?
Edit: Apologies for the bad tex