ε is preassigned, choosen or given in the (ε, δ)-definition of limit? Most of the textbooks as I have ever seen don't point out ε to be preassigned, choosen or  given  in the (ε, δ)-definition of limit, but some does, that is, the definition of limit becomes : For every preassigned, choosen or  given real  ε > 0, there exists a real δ > 0 such that for all real x, 0 < | x − a | < δ implies | f(x) − L | < ε.
So is there a need to obviously emphasize that ε is preassigned, choosen or  given  in the (ε, δ)-definition of limit?
 A: The real answer to your question is to be found in textbooks on the foundations of mathematics, which explain the meanings of the quantifiers "for all" and "there exists". These quantifiers always have the same meaning in mathematics. Here's a very brief overview (copied from my comments).
Suppose I have a mathematical sentence with a variable $\epsilon$, let's denote that sentence $P(\epsilon)$. The statement "for all $\epsilon>0, P(\epsilon)$" means that $P(.01)$ is true and $P(.0001)$ is true and $P(.00000000001)$ is true and $P(.0000000000409850473265)$ is true, and in fact no matter what positive number I subsitute for $\epsilon$, $P(\epsilon)$ is true. In order to prove that the sentence "for all $\epsilon>0, P(\epsilon)$" is true, I need to give an abstract mathematical argument which starts with the assumption $\epsilon > 0$ and ends with the proof that $P(\epsilon)$ is true.
Now suppose I have a mathematical sentence with a variable $\delta$, let's denote that sentence $Q(\delta)$. The statement "there exists $\delta > 0, Q(\delta)$" means that amongst all the many possible choice of a positive number $\delta$, somewhere there is one value of $\delta$ which makes $Q(\delta)$ true  (there may be more than one, but I don't care, all I need is one). It might not be $\delta = .1$, so I have no guarantee that $Q(.1)$ is true; it might not be $\delta = .0000004987$, so I have no guarnatee that $Q(.0000004987)$ is true; but there is some positive value of $\delta$ which makes $Q(\delta)$ true. In order to prove that the sentence "there exists $\delta > 0, Q(\delta)$" is true, I first need to find a particular value of $\delta$, perhaps by solving an equation, or solving an inequality, or digging it out at the bottom of some deep dark mine, or hunting it down in the middle of a deep dark jungle, bring that particular value of $\delta$ out and show it to everybody --- as you can see, this is the fun part of mathematics, especially if you like mining or hunting --- substitute that particular value of $\delta$ into the sentence $Q(\delta)$, and then give a mathematical argument that $Q(\delta)$ is true with that particular $\delta$ substituted.
Now let's put it together and suppose I have a statement $R(\epsilon,\delta)$ which has two variables in it. The statement "for all $\epsilon > 0$ there exists $\delta > 0$ such that $R(\epsilon,\delta)$" means that no matter what positive number I substitute for $\epsilon$, somewhere there is a value of $\delta$ which makes $R(\epsilon,\delta)$ true (there may be more than one, but I don't care, all I need is one). In order to prove that the statement "for all $\epsilon > 0$ there exists $\delta > 0$ such that $R(\epsilon,\delta)$", I need to give a mathematical argument which starts with the assumption that $\epsilon > 0$, and then I need to find a particular value of $\delta$ which depends on $\epsilon$ --- perhaps given by a particular formula $\delta = \delta(\epsilon)$ where the right hand side of the formula is a particular functional expression with $\epsilon$'s in it --- and then substitute that formula $\delta(\epsilon)$ into the sentence $R(\epsilon,\delta)$ which converts it into a new sentence having the form $R(\epsilon,\delta(\epsilon))$. Then I must give an abstract mathematical argument which starts with the assumption that $\epsilon>0$, and concludes by verifying that $R(\epsilon,\delta(\epsilon))$ is true.
A: A shorter answer is to reformulate the definition in the following shorter way: there exists a continuous positive strictly increasing function $\alpha(\epsilon)$ satisfying $\alpha(0)=0$ such that whenever the condition $0<|x-c|<\alpha(\epsilon)$ is satisfied, one necessarily has $|f(x)-f(c)|<\epsilon$. This lets one get out of the linguistic jungle of "preassigned, chosen or given" altogether. 
