How to choose $\epsilon$ for limit of a variable? 
A constant number $a$ is said to be the limit of a variable $x$, if
  for every preassigned arbitrarily small positive number $\epsilon$ it
  is possible to indicate a value of the variable $x$ such that all
  subsequent values of the variable will satisfy the inequality
  $|x-a|<\epsilon$

Does this mean that $\epsilon$ will change according to the problem under consideration?
For example, if the problem needs a variable $x$ in the range of $[0,2]$, then $\epsilon$ can be $0.1$ or $0.01$
but if the problem needs a variable $x$ in the range of $[0,2000]$, then $\epsilon$ can be $1$ or $2$
I am confused by the word "preassigned", can I just replace "every preassigned" with "any" ?
 A: It's attempting to highlight the fact that the interval of values satisfying $|x-a|<\epsilon$ doesn't depend on the value of $\epsilon$. I think it's a bad way to put things. You can think of this as "for every $\epsilon$ there exists an interval around $a$ such that...
Symbolically, we understand this to mean 
$\forall\epsilon\exists\delta$, where $\delta$ depends on $\epsilon$ such that $|x-a|<\delta\Rightarrow |x-a|<\epsilon$. Here $\delta$ gives the radius of the interval, because it's center is $a$. We can tell that this is true by picking $\delta=\epsilon/2$. Now we have that $$\forall\epsilon, |x-a|<\epsilon/2\Rightarrow |x-a|<\epsilon$$ which can is a true statement.
One reason this might be confusing is that as $x$ goes to $a$ the limit of $x$ is just $x$. I think your paragraph is exceptionally unclear. In general, we say that $L$ is the limit of $f(x)$ as $x$ goes to $a$ if 
$\forall\epsilon\exists\delta$, such that $|x-a|<\delta\Rightarrow |f(x)-a|<\epsilon$.
A: If $L$ is limit of $f$ at a point $a$.   
$\epsilon$ should be arbitrary, but $\delta$ you can choose it to satisfies the definition.
So that ,for every   $\epsilon >0$ there exists $\delta>0$ such that $0<|x-a|<\delta$ ,then $|f(x)-L|<\epsilon$
