Do these $\delta-\epsilon$ proofs work? 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

$$ \lim_{x\to -2} (2x^2+5x+3)= 1 $$
Finding $\delta$: $|2x^2 + 5x + 3 - 1| < \epsilon$
$|(2x + 1)(x + 2)| < \epsilon$
$|x - (-2)| < \delta $, pick $\delta = 3$
$|x+2| < 3 \Rightarrow -5 < x < 1 \Rightarrow -9 < 2x + 1 < 3$
This implies $|2x + 1||x + 2| < 3 \cdot |x + 2| < \epsilon \Rightarrow |x+2| < \frac{\epsilon}{3} $
$\\[22pt]$
Actual Proof: Let $\epsilon > 0 $. Choose $\delta = min\{3,\frac{\epsilon}{3}\}$
and assume that $0 < |x + 2| < \delta \Rightarrow |2x + 1||x+2| < 3 \cdot |x + 2| < 3 \cdot \frac{\epsilon}{3} = \epsilon$
