In negating logical statements, how does one decide what a logical connective "and" (=conjunction) is, and what a non-connective "and" is? I came across this problem while trying to work out the symbolic expression of a particular definition, and have been stumped since. 
Take this definition of continuity
A function $f\colon D \to R \,$  is continuous at $c \in D \,$ if for every $\epsilon >0$ there is a $\delta>0$ such that $|f(x)-f(c)|<\epsilon \,$ if $\, |x-c|<\delta$ and $x \in D$ .
However, in trying to express this in symbolic terms, I run into a problem. The implication part of this definition, i.e. "$|f(x)-f(c)|<\epsilon \,$ if $\, |x-c|<\delta$ and $x \in D \,$", looks to me like $ x \in D \ \land \ |x-c|<\delta \implies |f(x)-f(c)|<\epsilon$ in symbolic terms. This is considered wrong however, apparently the "and" in my definition isn't a conjunction, but just a non-connective kind of "and". What exactly is this "and" then, and how do I distinguish between logical connective and's and non-connective and's?
It's also worth noting that $x \in D$ in the implication should be translated to symbolic terms $\forall x \in D$, according to my book. Why? The resulting statement would be 
A function $f\colon D \to R \,$  is continuous at $c \in D \,$ if for every $\epsilon >0$ there is a $\delta>0$ such that if $\, |x-c|<\delta$ and for all $x \in D$, then $|f(x)-f(c)|<\epsilon \,$ 
...which sounds plain ridiculous and incomprehensible. 
Could anybody offer some insight?
 A: Anyone who claims that it is wrong to interpret this "and" as a conjunction is simply wrong.  There is absolutely nothing wrong with translating $$\text{"$|f(x)-f(c)|<\epsilon \,$ if $\, |x-c|<\delta$ and $x \in D \,$"}$$ as $$( x \in D \ \land \ |x-c|<\delta) \implies |f(x)-f(c)|<\epsilon$$ (though note the parentheses which you left out).  The meaning of this "and" is exactly the usual conjunctive meaning and it is not a different kind of "and".  Note, though, that in the context of the definition of continuity there is an implicit universal quantifier on the variable $x$ (the statement must be true for all $x$), so enlarging the context a little bit you could also translate it as $$\forall x(( x \in D \ \land \ |x-c|<\delta) \implies |f(x)-f(c)|<\epsilon).$$
When your book suggests to interpret it as $\forall x\in D$, they mean to put it at the start of the conditional, not in the middle as you did.  In doing so, they are rephrasing the statement as a logically equivalent statement which leaves out the "and" (and thus the conjunction).  Specifically, they mean $$\text{"for all $x\in D$, if $\, |x-c|<\delta$, then $|f(x)-f(c)|<\epsilon$"}$$ or in symbols, $$\forall x\in D\ (|x-c|<\delta \implies |f(x)-f(c)|<\epsilon).$$
Note that this statement is logically equivalent to your version with $\forall x$ at the start, so both are equally correct.
(As a final sidenote, there are uses of "and" which are not conjunctions.  For instance, if I say "the two largest prime numbers less than $50$ are $47$ and $43$" my "and" isn't really a conjunction, and I am using a construction of the English language that is fairly complicated to transform into formal logic.  This is totally unrelated to what is going on in your situation though.)
A: 
However, in trying to express this in symbolic terms, I run into a problem. The implication part of this definition, i.e. "$|f(x)−f(c)|<ϵ$ if $|x−c|<δ$ and $x∈D$", looks to me like $x∈D ∧ |x−c|<δ⟹|f(x)−f(c)|<ϵ$
  in symbolic terms. This is considered wrong however, 

It is not exactly wrong, but perhaps you should add parenthesis to be clear you have the correct form of $(A\wedge B)\to C$, rather than the completely different $A\wedge (B\to C)$.

apparently the "and" in my definition isn't a conjunction, but just a non-connective kind of "and". What exactly is this "and" then, and how do I distinguish between logical connective and's and non-connective and's?
  It's also worth noting that $x∈D$ in the implication should be translated to symbolic terms $∀x∈D$, according to my book. Why? 

There is an implicit universal quantifier, because $x$ is not a particular constant value, but rather an arbitrary variable.
$$\forall x~\big((x\in D \wedge \lvert x−c\rvert <\delta)~\to~(\lvert f(x)−f(c)\rvert <\epsilon)\big)$$
Since $(A\wedge B)\to C$ is equivalent to $A\to (B\to C)$, this can also be written as : 
$$\forall x~\Big(x\in D \to \big((\lvert x−c\rvert <\delta)~\to~(\lvert f(x)−f(c)\rvert <\epsilon)\big)\Big)$$
And by recalling that: $\forall x~(x\in D\to P(x))$ is often abbreviated as $\forall x{\in}D~P(x)$
$$\forall x{\in}D~\big((\lvert x−c\rvert <\delta)~\to~(\lvert f(x)−f(c)\rvert <\epsilon)\big)$$
