What are the logical underpinnings of the epsilon- delta definition of limits? I'm having trouble getting my head around the epsilon-delta defintion of limits. I learned about conditional statements and I know that in order for a conditional to be true , one of the following cases have to hold 


*

*both antecedent and consequent are true, or

*both are false, or

*the consequent is true and antecedent is false.


Now in proving the definition of a limit say
$$\lim_{x \rightarrow 4} x^2 + x -11 = 9 $$
which case do we prove? Is it the same for every limit or are different limits proved by different cases?
How does the concept of conditional truth-tables relate to this definition?
Thank you in advance.
 A: If you're proving that $P\implies Q$, for some statements $P$ and $Q$, you don't care which of these 
$$\underbrace{P\textsf{ true},\;Q\textsf{ true}}_{\large P\land Q}\qquad\qquad \underbrace{P\textsf{ false},\;Q\textsf{ true}}_{\large (\lnot P)\land Q}\qquad\qquad \underbrace{P\textsf{ false},\;Q\textsf{ false}}_{\large (\lnot P)\land(\lnot Q)}$$
is true. Your only task is to prove that
$$\underbrace{P\textsf{ true},\;Q\textsf{ false}}_{\large P\land(\lnot Q)}$$
is false. In other words, these statements are logically equivalent:
$$(P\implies Q)\iff\lnot(P\land(\lnot Q))$$

Given a function $f:\mathbb{R}\to\mathbb{R}$, and real numbers $a,L\in\mathbb{R}$, the meaning of the statement
$$\lim_{x\to a}f(x)=L$$
is that
$$\textsf{for any }\epsilon>0,\textsf{ there exists some }\delta>0\textsf{ such that for all }x\in\mathbb{R},\\
|x-a|<\delta\implies|f(x)-L|<\epsilon$$
Let's take a look at your example of
$$\lim_{x\to 4}x^2+x-11=9$$
so $\,a=4$, $\,L=9$, $\,f(x)=x^2+x-11$. If $\epsilon=1$, then you want to find a $\delta>0$ such that
$$\textsf{for all }x\in\mathbb{R},\;\;|x-4|<\delta\implies|x^2+x-11-9|<1$$
An example of such a $\delta$ would be $\delta=\frac{1}{10}$. In order to prove that
$$\textsf{for all }x\in\mathbb{R},\;\;|x-4|<\tfrac{1}{10}\implies|x^2+x-20|<1$$
all we care about is proving that there is no $x\in\mathbb{R}$ such that
$$|x-4|<\tfrac{1}{10}\textsf{ true},\;|x^2+x-20|<1\textsf{ false}$$
For example,
$$|5-4|<\tfrac{1}{10}\textsf{ false},\;|5^2+5-20|<1\textsf{ false}\qquad\quad \textsf{great! }\checkmark$$
$$|4.109-4|<\tfrac{1}{10}\textsf{ false},\;|(4.109)^2+(4.109)-20|<1\textsf{ true}\qquad\quad \textsf{great! }\checkmark$$
$$|4.05-4|<\tfrac{1}{10}\textsf{ true},\;|(4.05)^2+(4.05)-20|<1\textsf{ true}\qquad\quad \textsf{great! }\checkmark$$
A: This is a good question! 
Strictly speaking, there is a logical implication here. But once you understand that we only care about one case, your job becomes much easier (and less confusing). Why do we only care about one case?
Another way to look at the statement "$Q$, whenever $P$" (equivalently, "If $P$, then $Q$") is that the statement is true if


*

*Both $P$ and $Q$ are true, or if 

*$P$ (the antecedent) is false. 


The latter case, when the antecedent is false, we often say that the statement "$Q$, whenever $P$" is vacuously true (and I think of it as "true by a technicality", or "true by default"). It's similar to the statement "If I go to the store, I'll bring you something". You wouldn't consider me a liar if I don't go to the store (regardless of whether I've brought you anything), would you? It's essentially the same kind of "true by a technicality." If I don't go to the store, then you don't even need to ask "Have you brought me anything?" because I technically couldn't possibly have lied to you.

Aside: My thoughts are that the logic we use in calculus isn't really the same kind of logic that's done in a purely logical setting; at least, we don't have the same goals. We really don't care about implications with false antecedents because we're only interested in using implications (and properties defined with implications, like continuity) to deduce things. For that we need modus ponens, which requires our antecedents to be true.
For this reason, I think a truth table will generally be nearly useless in a calculus setting.

So, when confronted with the task of showing that

for all $\epsilon > 0$, there exists a $\delta >0$ such that $\lvert f(x) - 9 \rvert < \epsilon$ whenever $\lvert x - 4 \rvert < \delta$, 

we do a few things:


*

*Pick a generic $\epsilon > 0$.

*Find some $\delta > 0$ so that when we assume $\lvert x - 4 \rvert < \delta$, we can show that $\lvert f(x) - 9 \rvert < \epsilon$. 
Step 2. is where all the work happens. It usually involves using some unknown $\delta$, assuming $\lvert x - 4 \rvert < \delta$, and seeing what kind of relationship we can derive, algebraically, between $\epsilon$ and $\delta$. 
(Notice that I said "we assume $\lvert x - 4 \rvert < \delta$". This goes back to the fact that, if $\lvert x - 4 \rvert \nless \delta$, then our antecedent is false, and the "$Q$, whenever $P$" statement involving the two inequalities is true by default -- there's literally nothing to show, so we don't even bother talking about it.)
Once we have our $\delta$, actually showing that $\lvert f(x) - 9 \rvert < \epsilon$ using our $\delta$ and $x$ such that $\lvert x - 4 \rvert < \delta$ can happen almost as if by magic: We generally don't even need to say how we found such a $\delta$ (although we had to go through the algebra in step 2. to do so), only that we've found one, and shown that it works.
A: Just to make sure we are on the same page, let me write out that conditional:
For all $\epsilon > 0$ there exists a $\delta > 0$ such that for all $x$
$$
|x - 4| < \delta \implies |(x^2 + x - 11) - 9| < \epsilon\\
$$
Note that the conditional appears AFTER you choose $\epsilon$, $\delta$, and $x$. For any choice of $\epsilon$, there needs to be some choice for $\delta$, such that for any choice of $x$, the conditional is true.
Which of those three cases holds will depend on the $\epsilon$, $\delta$ and $x$ you choose!
Say you take a small $\epsilon$, choose a $\delta$ small enough for the $\epsilon$, and take an $x$ that is very far from 4 (like 100 or something). Then:
$$
|x - 4| > \delta
$$
$$
|(x^2 + x - 11) - 9| > \epsilon
$$
So the antecedant is false, the consequent is false, and the conditional is true.
Now say you take an $x$ that is very far from 4, but very close to -5. Then:
$$
|x - 4| > \delta
$$
$$
|(x^2 + x - 11) - 9| < \epsilon
$$
(Because -5 is the other solution to the equation $x^2 + x - 11 = 9$.)
So the antecedant is false, the consequent is true, and the conditional is true.
Finally, say you take an $x$ that is very close to 4. Then:
$$
|x - 4| < \delta
$$
$$
|(x^2 + x - 11) - 9| < \epsilon
$$
So the antecedant and consequent are both true, and the conditional is true.
So the answer is: you use all three cases for every problem! Which one holds depends on which $x$ you are looking at.
