Why does this proof invoke the Archimedean principle? Okay so I am having trouble understanding this proof. What is really bugging me is how the Archimedean principle is referenced. How does one know that this is what is needed to proceed and how is it even relevant? I do not see the significance of N > 1/epsilon. Can someone please explain? Thank you.

 A: Well, to prove that $1/n\to 0$, you need to prove that for any $\epsilon>0$ there exists $N$ such that if $n\geq N$ then $|1/n|<\epsilon$ (this is just the definition of a limit).  Since $1/n$ is positive, this just means you have to prove that $1/n<\epsilon$ for all sufficiently large $n$.  Taking reciprocals, this is equivalent to $n>1/\epsilon$.
So in order to prove that $1/n\to 0$, you exactly need to know that if $n$ is a sufficiently large natural number, then $n>1/\epsilon$ (for any fixed positive $\epsilon$).  This is exactly what the Archimedean principle says (every real number is less than some natural number), so that's why you invoke it.
A: So, the Archimedean property essentially uses the fact that $\mathbb N \subseteq \mathbb R$ is an unbounded subset.
If for each $r \in \mathbb R$, there is a natural number $n$ so that $n-1 \leq r <n$, which is equivalent to asserting that $1/n<1/r$.
To see the statement:
First, notice that $(1/n)_{n \in \mathbb N}$ is a decreasing sequence, so it will suffice to show that there exists some $N \in \mathbb N$ large enough so that $1/N<\epsilon$ which is equivalent to finding $N>1/\epsilon$. This can be done, since the natural numbers are unbounded. 
This allows us to say that $(1/n) \to 0$, since for our choice of $\epsilon>0$, we can find some $N \in \mathbb N$ so that $1/k<\epsilon$ for all $k \geq N$.
