How to use $\epsilon$-$\delta$ definition to prove that $\lim_{x\to7}(x^3-9x^2-x+15) = -90$ I know that $\epsilon$ > 0, and that $\delta$ > 0
I have found this post which deals to a similar practice problem I am having, but the thing is my problem uses 3 roots, and I have no idea what to do after this point.
$$\lim_{x\to7}\left(x^3-9x^2-x+15\right)\ = -90$$
Here you can get three roots, which are |(x-5)(x-7)(x+3)| < $\epsilon$
In general, my main question is how do I know which 2 roots to use and why?
Proving a limit using epsilon delta definition
 A: $$|(x-5)(x-7)(x+3)|\leq  (|x-7|+2)|x-7| (|x-7|+10)$$ So $|x-7| <\delta$ gives $$|(x-5)(x-7)(x+3)| <(\delta+2)\delta (\delta+10)$$ In particular, if $\delta <1$ this gives $$|(x-5)(x-7)(x+3)| <(3)(\delta) (11)$$ so it is enough to take $\delta <1$ and $\delta <\frac {\epsilon} {33}$.
A: You need to use all the roots. Here I will use the method outlined in the most upvoted answer (because it is indeed the most straightforward).
First suppose $|x-7| <\delta$. Furthermore we restrict $\delta < 1$, so we only consider $6 < x < 8$.
Using the factorization $|x^3 - 9x^2-x+15+90| = |x-5||x-7|||x+3|$, we still need to bound $|x-5|$ and $|x+3|$. But with the restriction $6 < x < 8$, we easily see that $|x-5| < 3$ and $|x+3| < 11$.
Now we end up with
$$|x^3 - 9x^2-x+15+90| = |x-5||x-7|||x+3| < 3 (\delta)(11) = 33\delta$$
So for any $\epsilon$, we can choose $\delta = \min\{1, \dfrac\epsilon {33}\}$.
A: You do not need a factorisation. Let $\epsilon>0$. For $|x-7|<\delta\le 1$, we have $|x|<8$, so $$|x^3-7^3| =|x-7||x^2+7x+7^2|\le|x-7|\underbrace{(8^2 + 7\times 8 + 49)}_{=:C_1}\le C_1 \delta,\\
|-9x^2 + 9\times 7^2| = 9|x+7||x-7| \le \underbrace{9 \times (8+7)}_{=:C_2} |x-7| \le C_2 \delta, $$
so if
$|x-7|\le \delta:=\frac{\min(\epsilon,1)}{C_1 + C_2  + 1}\le 1$, then as $-90 = 7^3 -9\times 7^2 -7 + 15,   $
\begin{align} |x^3 - 9x^2 -x+15 - (-90)| &\le |x^3-7^3| + |-9x+9\times 7^2|  + |-x+7| \\
&\le \frac{C_1\min(\epsilon,1)}{C_1+C_2+1} + \frac{C_2\min(\epsilon,1)}{C_1+C_2+1} + \frac{\min(\epsilon,1)}{C_1+C_2+1}
\\
&=\min(\epsilon,1)\\ &\le \epsilon.
\end{align}
This is easier to replicate when  factorisation is hard e.g. the degree of the polynomial is large, or you're checking the limit at 'unusual' points. (This is just product rule and sum rule repeatedly applied.)
