Prove $\lim \limits_{x \to 3}{(x^2+x-4)} = 8$ via the precise definition of a limit. Exercise

Prove the statement using the $\epsilon$, $\delta$ definition of a limit:
  $$\lim \limits_{x \to 3}{(x^2+x-4)} = 8$$


The Precise Definition of a Limit
In case you're not familiar with the definition of "The Precise Definition of a Limit", here it is.

Let $f$ be a function defined on some open interval that contains the number $a$, except possible $a$ itself. Then we say that the limit of $f(x)$ as $x$ approaches $a$ is $L$, and we write
  $$\lim \limits_{x \to a}{f(x)} = L$$
  if for every number $\epsilon > 0$ there is a number $\delta > 0$ such that
  $$\text{if } 0 < |x - a| < \delta \text{ then } |f(x) - L| < \epsilon$$


Attempt
$\lim \limits_{x \to 3}{(x^2+x-4)} = 8 \implies \text{if } 0 < |x - 3| < \delta \text{ then } |x^2+x-4 - 8| < \epsilon$
$|x^2+x-4 - 8| < \epsilon \implies |x^2+x-12| < \epsilon \implies |(x-3)(x+4)| < \epsilon$
From here, I quickly get lost.
I notice that $|(x-3)(x+4)| < \epsilon \implies |x-3||x+4| < \epsilon \implies |x-3| < \frac{\epsilon}{|x+4|}$, which means that $\delta = \frac{\epsilon}{|x+4|}$.
In the other exercises, I often end the proof here, because I've solved for $\delta$ purely in terms of $\epsilon$. However, here I have $x$ on the RHS.
 A: Since you're looking at the limit as $x\to3$ one could assume that as $x$ approaches this value, it is within a distance of say $1$ from $3$. That is, $|x-3|<1$ for sufficiently close $x$ (which is the case since $x\to3$). This inequality gives:
$$|x-3|<1\\2<x<4\\6<x+4<8$$
Thus $|x+4|$ is bounded between $6$ and $8$. Returning to what you had: $|x-3|<\frac{\epsilon}{|x+4|}$, we see that the minimum of $\frac{\epsilon}{|x+4|}$ occurs when $|x+4|$ is maximized, which is $8$ (as we showed). Thus$|x-3||x+4|<\epsilon$ so $|x-3|<\frac{\epsilon}{8}$. Now we have two inequalities:
$$|x-3|<1$$
and $$|x-3|<\frac{\epsilon}{8}$$
So try choosing $\delta$ as the minimum of $1$ and $\frac{\epsilon}{8}$, and verify that it works.
A: Hint: write $f(x)=x^2+x-4=(x-3)^2+7(x-3)+8$ and let $\epsilon \gt 0 $. Then:
$$|f(x)-8| \lt \epsilon \;\;\iff\;\; |(x-3)^2 + 7(x-3)| \lt \epsilon$$
Choose $\delta = \min(1, \epsilon / 8)$ and let $0 \lt |x-3| \lt \delta$. Then:


*

*$0 \lt |x-3| \lt \delta \le 1$ implies that $|x-3|^2 \lt |x-3|$

*$|x-3| \lt \delta \le \epsilon / 8$ implies that $8|x-3| \lt \epsilon$ 
Using the above and the triangle inequality:
$$|(x-3)^2 + 7(x-3)| \le |x-3|^2 + 7|x-3| \lt |x-3| + 7|x-3| = 8 |x-3| \lt \epsilon$$
A: Given $\epsilon>0$, choose $\delta=\min\{1,\epsilon/8\}$, then $-1<x-3<1$ implies that $2<x<4$. Thus, $6<x+4=|x+4|<8$ and hence to have $|x^2+x-12|<\epsilon$, you need to choose $\delta$ such that $|x-3|<\epsilon/8$.
