Question about a worked out problem regarding epsilon-delta Proof my question is regarding the last part since $|x+3|<6$ I thought it must be that $$
|x-2|<\frac{\epsilon}{6}<\frac{\epsilon}{|x+3|} 
$$
and not $$
|x-2|<\frac{\epsilon}{|x+3|}<\frac{\epsilon}{6}
$$ I think I'm missing something, maybe someone can help me with this problem. the whole proof is in this paper it is the 4th proof.
$$
\lim _{x \rightarrow 2} x^{2}+x-2=4
$$
We start the same way we always do. By now this part should be automatic.
$$
\begin{aligned}
|f(x)-L|<\epsilon & \Longrightarrow\left|\left(x^{2}+x-2\right)-4\right|<\epsilon \\
& \Longrightarrow\left|\left(x^{2}+x-6\right)\right|<\epsilon \\
& \Longrightarrow|(x+3)(x-2)|<\epsilon \\
& \Longrightarrow|x+3||x-2|<\epsilon \\
& \Longrightarrow|x-2|<\frac{\epsilon}{|x+3|}
\end{aligned}
$$
Once again we have the left side in the form $|x-a| .$ Now we can let $\delta$ equal $\frac{\epsilon}{|x+3|}$. But what do we do about the $|x+3| .$ In general $\delta$ must be in terms of $\epsilon$ only, without any extra variables.
So how we can remove this $x+3$ term?
First we need to simplify the problem a little bit. Since the concept of limit only applies when $x$ is close to $a$, we will first restrict $x$ so that it is at most 1 away from $a$, or, mathematically, $|x-a|<1$ (in our case $|x-2|<1$. Then, this means, $1<x<3$, or $4<x+3<6$.
Now consider the original inequality
$$
\begin{array}{c}
|x-2|<\frac{\epsilon}{|x+3|} 
\end{array}
$$
Notice that the right hand side is at the minimum when $x+3$ is at its maximum. Since the maximum of $x+3$ is $6,$ we know that
$$
|x-2|<\frac{\epsilon}{|x+3|}<\frac{\epsilon}{6}
$$
 A: It seems that your logic is going in the wrong direction. While it is often best to start with $|f(x) - L| < \varepsilon$ when figuring out the proof, ultimately your logical argument has to start with assuming $0 < |x - a| < \delta$ (where $\delta$ is the function of $\varepsilon$ you've chosen), and it has to end by concluding $|f(x) - L| < \varepsilon$.
Using your work so far, it looks like we need $\delta = \min\{1, \varepsilon / 6\}$. We then have
\begin{align*}
0 < |x - 2| < \min\left\{1, \frac{\varepsilon}{6}\right\} &\implies |x - 2| < 1 \text{ and } 6|x - 2| < \varepsilon \\
&\implies |x + 3| < 6 \text{ and } 6|x - 2| < \varepsilon \\
&\implies |x + 3||x - 2| < 6|x - 2| < \varepsilon \\
&\implies |x^2 + x - 2 - 4| < \varepsilon,
\end{align*}
completing the proof. Note that the logic makes perfect sense, when written in order!
EDIT: Let me see if I can explain a bit better. I still think this is a misconception about the direction of the flow of logic, but we can make this clearer, I think. You are absolutely right when you say it should be
$$|x-2|<\frac{\varepsilon}{6}<\frac{\varepsilon}{|x+3|}, \tag{1}$$
not
$$|x-2|<\frac{\varepsilon}{|x+3|}<\frac{\varepsilon}{6}. \tag{2}$$
And it is. Really, we are using $(1)$, not $(2)$. We want to put the step $|x - 2| < \varepsilon/6$ (where the $\varepsilon/6$ is derived from our choice of $\delta$) before the step $|x - 2| < \varepsilon / |x + 3|$. We want the logic to go:
$$|x - 2| < \frac{\varepsilon}{6} \implies |x - 2| < \frac{\varepsilon}{|x+3|},$$
which is what $(1)$, not $(2)$, delivers to us. This is because, for our choice of $\delta$, $|x - 2| < \varepsilon/6$ follows easily from $0 < |x - 2| < \delta$, and $|x^2 + x - 2 - 4| < \varepsilon$ follows easily from $|x - 2| < \varepsilon / |x + 3|$.
As I said, I think you were viewing this from the wrong direction. In your working, you very explicitly started with $|x^2 + x - 2 - 4| < \varepsilon$, and worked backwards. You reached the step $|x - 2| < \varepsilon / |x + 3|$ first, and started looking for ways to show that this implied $|x - 2| < \varepsilon/6$.
This implication is false. In order for something like that to be true, you'd want $(2)$ to be true, but it isn't. That way, you could then conclude that $|x - 2| < \varepsilon / 6$, and then end with $|x - 2| < \delta$. But this is simply not what the definition of a limit requires. You don't need to show
$$0 < |x - a| < \delta \impliedby |f(x) - L| < \varepsilon,$$
you need to show the converse instead.
