How to prove that a limit exists for $\lim_\limits{n\to \infty}\frac{n+2}{n} = 1$ i know that if I want to prove that $\lim_\limits{n\to \infty}\frac{n+2}{n} = 1$, I have to use the Epsilon and Delta values, as following:
$$\left|\dfrac{n+2}{n}-1\right|<ε$$ 
$$\left|\dfrac{n+2-n}{n}\right|<ε$$
$$\left|\dfrac{2}{n}\right|<ε$$
$$\left|\dfrac{2}{ε}\right|<n$$
Then I assume that, $ = \left|\dfrac{2}{ε}\right|$, but I'm not sure how to proceed from here. Also the method in my lecture was a different to the one you see in mainly all explanations of limits. My professor used $$\frac{1}{n}≤\frac{1}{N_ε}≤\frac{1}{1/n}=ε$$ to explain the idea of limits, which has me insanely lost.
 A: $a_n = \dfrac{n+2}{n};$
$|a_n -1| =\dfrac{2}{n} .$
Let $\epsilon >0$ be given.
Choose  $M \ge 2/\epsilon$, $M $, real. positive.
There is a $n_0 \in \mathbb{N}$ such that $n_0 >  M.$
(Archimedes)
For $n\ge n_0 ;$
$|a_n-1| = 2/n \lt 2/n_0 \lt 2/M \le \epsilon, $ 
I.e. $\lim_{n \rightarrow \infty} a_n =1$.
A: You want: given any distance $\epsilon>0$, there is some $N_\epsilon$ (depending on $\epsilon$, hence the subscript) such that for all $n\geq N_\epsilon$, $\frac{n+2}{n}$ is within distance $\epsilon$ with the proposed limit $1$.
You've shown: $|\frac{n+2}{n}-1|=|\frac{2}{n}|=\frac{2}{n}$. And if $n\geq N_\epsilon$, then $\frac{2}{n}\leq\frac{2}{N_\epsilon}$. So to ensure $\frac{2}{n}<\epsilon$, it is enough to ensure $\frac{2}{N_\epsilon}<\epsilon$ which is equivalent to $N_\epsilon>\frac{2}{\epsilon}$. So, for example, pick $N_\epsilon=1+\lceil\frac{2}{\epsilon}\rceil$.
A: You will arrive at something like what your professor wrote to you after you have sufficient amount of experiences with the epsilon-analysis. For the current problem, just note that 
$$
|\frac{n+2}{n}-1| = \frac{2}{n}
$$
for all $n$; given any $\varepsilon > 0$, we have $2/n < \varepsilon$ if $n > 2/\varepsilon$. So taking $N := 1+ \lceil 2/\varepsilon \rceil$, i.e. $1$ plus the smallest integer $\geq 2/\varepsilon$, suffices.
