proof of a limit with epsilon - N definition for sequences I decided to do something useful with time during the lockdown due to Coronavirus and started learning calculus.
I recently got to the definition of a limit of a sequence, and I have a problem understanding how epsilon and $N$ are determined/decided upon.
For example, the professor solved this proof of $1/n$ is $0$ as $n$ approaches $\infty$ in the following way:
He started solving by $|1/n-0|<\epsilon$, $1/n <\epsilon$ and then the professor defined $N$ to be $(1/n)<(1/N)<\epsilon$.
I don't understand why he defined $N$ like that, and how I should define  $N$ in terms of epsilon.
(if my question is not clear enough, please do not close this question, I will edit it to your demands)
Thank you very much! 
 A: The idea of $\epsilon$-$\delta$ proofs is that you think of $\epsilon$ as an arbitrarily small quantity in which "after" $\delta$, the sequence is $\epsilon$ close to the limit of the sequence.
Students first studying $\epsilon$-$\delta$ proofs are usually told to first "sketch" their proof and then write it out formally. Your professor has done the latter to show you how one arrives at $\delta$ in terms of $\epsilon$. 
Perhaps a good way for you to understand this is following this slightly more concrete example. Let $f(n) = \frac{1}{n}$ be a sequence, we know that $f(n)$ approaches $0$ as $n \to \infty$. Now, consider $\epsilon = 0.1$. When is $|f(n)| < 0.1$?. We have that $\frac{1}{9} > 0.1$ so that doesn't work. $\frac{1}{10} = 0.1$, close but we are not looking for an equality, we are looking for a strict inequality. $\frac{1}{11}$ works and so does $\frac{1}{12}, \frac{1}{13},...$. This can be written out more formally as: for all $n > 10$, $\frac{1}{n} < 0.1$. 
$N$ is that "magic number" in which considering any number strictly higher than $N$ yields the difference of that function and the limit less than epsilon. In our example, $N = 10$. But notice that we could've chosen $N = 11, 12, ...$ instead and it would've worked the same.
The objective of $\epsilon$-$\delta$ is to then consider any arbitrary $\epsilon > 0$ and then find a $\delta$ depending on $\epsilon$ that will satisfy the desired property. 
An analogy my first year calculus professor told us is to think of it as a war between $\epsilon$ and $\delta$. It goes like this: $\epsilon$ wins by throwing a small enough number so that the function is never closer than that number to the limit. $\delta$ wins if for any number $\epsilon$ throws, he/she can always respond back with a high number so that the function evaluated at numbers higher than that one is $\epsilon$ close to the limit.
A: 
Definition of limit:
  $\lim\limits_{n\to \infty}x_n=L$ if and only if
  For any $\epsilon>0$, there exists $N$(which may depends on the $\epsilon$) such that $|x_n-L|<\epsilon$ whenever $n>N$

This means that for any positive value of $\epsilon$ that we choose, the distance between $x_n$ and $L$ will be smaller than $\epsilon$ as long as $n$ is large. (Remember that we always choose the value of $\epsilon$ first and determine the value of N afterwards.)
Following your example of $1/n$, we start with an arbitrary $\epsilon>0$.
Let's say we choose $\epsilon=0.01 >0$
We want the difference $|x_n-0|=|\frac{1}{n}|$ to be smaller than $0.01$ whenever $n>N$ for some $N$.
 In this case, if we choose $N=100$, then we can ensure that $|\frac{1}{n}|<0.01$
when $n>N$.
Of course the above is just a special case for $\epsilon$, but you can easily generalize the above statement to make it work for any $\epsilon>0$
