<s>[INST] Absolute Value within an Epsilon Delta Problem I'm having trouble with an epsilon-delta proof. I'm asked to prove the limit as $x$ goes to $-2$ of $f(x)=|x^2-9|/(x^2+3x+1)$ is $-5$, so we have $|(|x^2-9|/(x^2+3x+1))-5|<ϵ$. I've seen that $|x^2-9|$ is $|x-3||x+3|$ but I'm not really sure how to proceed to 
"obtain" the desired $|x-a|=|x+2|<δ$. Am I supposed to place a limitation on either $|x-3|$ or $|x+3|$? Or is there some other way to handle this problem? Thanks in advance.
 [/INST] Using the advice provided by @Robert Shore, we can get this
$$
\vert x^2-9\vert = \vert 9 - x^2 \vert = 9-x^2 \iff x^2\leq9 \implies -3\leq x\leq 3
$$
since we are checking the limit as $x\rightarrow2$, restrict $\delta < \frac{1}{2}$. Now
$$
\vert\dfrac{\vert 9 - x^2 \vert}{x^2+3x+1} + 5\vert < \varepsilon
$$
we can modify this by bringing everything to a common denominator
$$
\vert\dfrac{ 9 - x^2 + 5(x^2+3x+1)}{x^2+3x+1}\vert = \vert\dfrac{4x^2+15x+14}{x^2+3x+1}\vert = \vert\dfrac{(4x+7)(x+2)}{x^2+3x+1}\vert
$$
now if $\delta<\frac{1}{2}$, then we can know that $|x+2|<\frac{1}{2} \implies -2.5<x<-1.5$. It turns out that the bottom polynomial has a root at $x\approx -2.6$ which is why we had to restrict $\delta<\frac{1}{2}$ instead of $1$
Now that we have a range of $x$, lets maximize the Left Hand Side by maximizing the numerator and minimizing the denominator.
$|4x+7|$ has a maximum on the interval at $x=-2.5$ which is equal to $3$
$|x^2+3x+1|$ has a minima at $x=-2.5$ equal to $\frac{1}{4}$ so we can say
$$
\vert\dfrac{(4x+7)(x+2)}{x^2+3x+1}\vert < |\dfrac{3(x+2)}{\frac{1}{4}}| = |12(x+2)| < \varepsilon \implies|x+2|<\dfrac{\varepsilon}{12}
$$
Now 
$$
\forall \varepsilon > 0, \exists \delta = \min\left(\frac{1}{2}, \frac{\varepsilon}{12}\right) \vert |x+2|<\delta \implies  |f(x)+5|<\varepsilon
$$
Q.E.D.
</s>