Prove $\lim_{n\to\infty}\frac{2n}{n+2}=2$ Prove $$\lim_{n\to\infty} \frac{2n}{n+2} = 2$$
I know for all $\epsilon > 0$ there exists some $k\in\mathbb{N}$ for all $n > K$ and $\left|\frac{2n}{n+2} - 0\right| < \epsilon$.
Proof:
Fix any $\epsilon > 0$ choose $k \in \mathbb{N}$ where there exists $???? < K$
I am getting stuck on what to choose for $K$ in terms of $\epsilon$
I know that $\frac{2n}{n+2} < \frac{2n}{n} = 2$ but from there I get stuck.

$\textbf{Proof:}$
Let $\epsilon >0$ be given.
$\textbf{RTP:}$ $\exists N_\epsilon \in \mathbb{N} \ni |\frac{2n}{n+2} -2|<\epsilon \ \forall \ n \geq N_\epsilon $.

$\textbf{ROUGH WORK:}$
We know that $|\frac{2n}{n+2}-2| = |\frac{-4}{n+2}| =\frac{4}{n+2} \forall  n\in \mathbb{N}$.
Now we need $\frac{4}{n+2} < \epsilon \implies \frac{n+2}{4} > \frac{1}{\epsilon}$ (Since all values are positive)
$\implies n+2 > \frac{4}{\epsilon} \implies n> \frac{4}{\epsilon}-2$.

For the given $\epsilon > 0$, by the Archimedean Property, we may choose $N_\epsilon > \frac{4}{\epsilon}-2$. 
We now have, $\forall n \geq N_\epsilon$ :
\begin{align*} n \geq N_\epsilon > \frac{4}{\epsilon} -2\end{align*}
$ \implies n+2 \geq N_\epsilon +2 > \frac{4}{\epsilon} \\ \implies \frac{n+2}{4} \geq \frac{N_\epsilon +2}{4} > \frac{1}{\epsilon} \\ \implies \frac{4}{n+2} \leq \frac{4}{N_\epsilon +2}<\epsilon \\ \implies \frac{4}{n+2}<\epsilon$
And hence, $\lim_{n\to \infty}(\frac{2n}{n+2})=2$
