An ABC soft question about epsilon-delta argument Someone told me that some textbooks present epsilon-delta argument somewhat misleadingly. For example, consider the simplest one: the convergence of the sequence $(1/n)_{1}^{\infty}$
to $0$.
These textbooks may prove this convergence as follows: Let $\epsilon > 0$. Since $n \geq N,$ we have $1/n \leq 1/N,$ so that if we choose $N := [1/\epsilon] + 1$ (where [x] denotes the greatest integer not greater than the given number $x$) then $n \geq N$ implies $1/n < \epsilon$.
Yes, such proof looks to prove the implication $n \geq N \implies 1/n < \epsilon.$ But indeed this implication is assumed valid and what requires to prove is the existence of $N$ for every $\epsilon > 0.$ 
I do not know how to reply to such question, would anyone please help? 

I think these textbooks are trying to lead you through how to get to the solution rather than set out a formal proof.
When I was taught epsilon-delta proofs we were told to figure out a suitable N in rough and then write out the argument formally.
So If we were told to prove $1/n \rightarrow 0$ as $n \rightarrow \infty$ then in rough I would think about how $n \ge N \implies 1/n \le 1/N $ and how I can use it to prove the limit.
Then I would state what we're trying to show:
$$\forall \epsilon > 0 \ \ \exists N \in \mathbb{N} \ \ n \ge N \implies 1/n < \epsilon $$
and then as my proof I would write -
"pick $N \in \mathbb{N}^{>0} $ such that $1/N<\epsilon . \ \ \ $ Then $\forall n \ge N \ \ \ $ $1/n \le 1/N < \epsilon $"
