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.