proof of limit explanation the teacher solved an exercise for example, and I don't understand why the solution is right.
I have been trying to understand the solution for hours. unfortuently I still don't understand.
proof that:
$$\lim\limits_{ n \to \infty}\frac{1}{n}=0$$
solution:
we need to proof that, for every epsilon, there is a N that for every an > N :
$$| \frac{1}{n}-0| < \varepsilon$$ ( I understood that in order to proof that 0 is the limit, I have to proof that the limit definition is true about this exercise )
$ \frac{1}{n} $ is always positive, because n is always positive, so: 
$ | \frac{1}{n} | = \frac{1}{n} $
$  \frac{1}{n} < \varepsilon $
$ n > \frac{1}{\varepsilon} $
( this part I didn't understand. what does it mean n > expression ? why do I have to get to " n > expression" in every lim proof? [that's what our teacher said])
The teacher wrote this example, here I understood nothing:
$ \varepsilon =\frac{1}{1000} $
so
n > 1000
thus,$  N = \frac{1}{\varepsilon} $
please help me
 A: Let $x_n=1/n$ then
$$\left|x_n-0\right|=\frac{1}{n}<\varepsilon\tag{*}$$
for any $\varepsilon>0$. We need to find a suitable $n$ that this holds for any arbitrarily small $\varepsilon$ and we get it via
$$\frac{1}{n}<\varepsilon\leftrightarrow n>\frac{1}{\varepsilon}.$$
This basically means that you can give me any accuracy $\varepsilon$ and I can find a $n$ s.t. $x_n<\varepsilon$. This is true for all $\varepsilon>0$. For example you pick $\varepsilon=0.5$ then I can see that I have to chose a $n>1/0.5=2$ which means any $n\geq 3$ would satisfy the equation $(*)$.
A: What you need to understand first is that $\varepsilon $ is a fixed positive number which is intuitively assumed to be very small. So by proving the statement for $\varepsilon$ without giving any initial value to it, you in fact prove it for any positive number that is fixed in advance.
This leads to the inequality:
$ n > \frac{1}{\varepsilon} $.
For better understanding, take an initial value like your teacher did, $\varepsilon=1/1000 $. This will lead you $ n > \frac{1}{1/1000} $, i.e. $ n > 1000 $. From here take as $N=1000$ beacause for any $n>N=1000$ the statement will be true for the given $\varepsilon=1/1000$ 
A: Essentially, by the limit you (try to) capture all numbers bigger than $n$ by an arbitrarily small $\varepsilon$. The proof is there to show that your capturing effort (by given $\varepsilon$) really works for all numbers bigger than $n$. 
By showing this, you can show that you can drive the $\varepsilon$ to any imaginable proximity to $0$ even to such distance, that $\varepsilon$ and $0$ become indistiguishable i.e. that the limit is equal to $0$ and still there will be an $n$ for which the above capture works up to infinity.
