Why $N$ is not the function of $\varepsilon$? Here is a calculus question:
$N$ is not the function of $\varepsilon$ in $\varepsilon-N$ definition.

I cannot understand why? In my opinion, every time we try to find a suitable $N$ which based on $\varepsilon$, that is, $N=N(\varepsilon)$. Why it is not a function? 

Thanks!
 A: Let me give you an example:
Prove: $$\lim_{n\to\infty}\frac{1}{n}=0$$
Proof: This is easy to see that we have to find a $N$, for every $n>N$ it satisfies 
$$|\frac{1}{n}-0|<\varepsilon\Rightarrow n>\frac{1}{\varepsilon}$$
Thus we can choose $N=[\frac{1}{\varepsilon}]+1$, while this is not the ONLY choice, which means you can also choose $N=[\frac{1}{\varepsilon}]+2,\ N=[\frac{1}{\varepsilon}]+3$, etc. Each of them will satisfy $\varepsilon-N$ definition. 
As you mentioned that "$N$ is based on $\varepsilon$", correct! But $N$ is not the ONLY one value which means it is not a function of $\varepsilon$ but we can say it is an expression of $\varepsilon$.
Hope this is helpful. 
A: May I say that I don't completely agree? Not to mention that there are also multivalued functions, there is nothing really wrong in thinking of $N$ as a function of $\varepsilon$. For instance, given $\varepsilon>0$, there is always a smallest $N$ as in the definition of limits, since any part of $\mathbb{N}$ has a minimum. 
Of course this is not the only way to define $N(\varepsilon)$, but I believe that such subtleties are really dangerous in a calculus course.
