How to explain Paul's guesswork as to some delta value in the context of a preliminary work for limit proof ( Paul's Online Math Notes). 
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*I highly valuate Paul's Online MathNotes, but, in spite of the efforts of the author, I can't manage to understand the guesswork he is doing while trying to determine some delta value in order to prove that : 


lim ( $x^2+x+11$) as x approaches 4 is equal to 9.
(Reference : Paul's Online MathNotes, Calculus I, Section 2-10, Definition of the limit, Example 3 http://tutorial.math.lamar.edu/Classes/CalcI/DefnOfLimit.aspx) 


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*In the example given in Paul's Online Math Notes ( see image below) , I do not understand why the author seems to give up his initial goal, which was to find a delta sufficient condition for "| x - 4| | x + 5| is less than  epsilon" , and "decides" to substitute for it the new goal : " K | x - 4 | is less than epsilon".
My question is not precise, and I am sorry for that. The reason is that, although having thought on this problem for more that an hour, my ideas are a total mess ( in spite of the didactic efforts of the author). 


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*What I am looking for is the logical structure of the author's heuristic work . How to present clearly the hierarchical structure of hypotheses and conclusions? What holds categorically? What hypothetically? Is he aiming at proving conditional statements? How can he hope that his assumptions will not make the whole results he's going to get conditionally valid ones? 

*I only understand some general principles about "delta guessing" 
(1) unlike the limit proof itself, it is a heuritic reasoning, not aiming at being deductively valid 
(2) it consists in starting with the consequent of the limit definition ( the " epsilon part" ) and trying to find some sufficient condition ( in terms of delta value) that makes this consequent true. 

 A: The "idea" (which is totally formalizable, but it's better to not try to think in definitions, but rather to think in ideas, and then write down the proof more strictly) is as follows.
We have the function $f(x)=x^2+x-11$. We want to know that we can get the value $$|f(x)-9|$$ as close to $0$ as we want by restricting $x$ to some interval around $4$.
More strictly: let's say that we are given some value of $\varepsilon$. We want to be sure that no matter what value we get, we can find some value of $\delta$ such that restricting $x$ to $(4-\delta, 4+\delta)$ will restrict $f(x)-9$ to $(-\varepsilon, \varepsilon)$.

OK, so, in the grand mathematical tradition... let $\varepsilon > 0$. 
No, wait, not yet. Let's first simplify $f(x)-9$. As you saw, $$|f(x)-9|=|x+5||x-4|.$$
How does this help us make this value small? Well, we have control over $\delta$, which means we can control how big the number $|x-4|$ will be, right? In fact, we know that $|x-4|<\delta$, whatever the $\delta$ we choose in future will be.
OK, how about the $|x+5|$ factor? Is it possible that this factor gets really really big? No, right? I mean, we can "promise", for example, that we will pick $\delta <1$, and in such a case, we know that $|x+5|\in (8, 10)$. So let's do that. Let's promise that whatever our $\delta$ will be, we promise it will be smaller than $1$.
In this case, we can see that we will get $$|f(x)-9|=|x+5||x-4| \leq 10\cdot \delta.$$
Note that there is nothing heuristic in what I wrote. It is a 100% certain deductively proven fact that if $\delta < 1$, then $|f(x)-9|\leq 10\delta$.

OK, so now what? Well, what we really want is $|f(x)-9|\leq \varepsilon$. Well, obviously, if we pick a $\delta$ such that $10\delta<\varepsilon$, then we are done, right? Because we now know that if $\delta < 1$ and $10\delta\varepsilon$, then $$|f(x)-9|\leq 10\delta < \varepsilon.$$
So, does there exist some $\delta$ that satisfies the conditions $\delta < 1$ and $10\delta<\varepsilon$?
Indeed it does! Many of them, in fact! But we only really need one, so let's pick it: $$\delta_0 = \min\left(1, \frac{\varepsilon}{20}\right).$$
Putting all that we wrote together, we know that for $\delta_0$, we know the following is true:

If $|x-4|<\delta_0$, then $|f(x)-9|<\varepsilon$.

But since $\varepsilon$ was arbitrary, our proof is complete.
