1. **Problem statement:** Prove that $$\lim_{n \to \infty} \frac{2n}{2+n} = 2$$ using the definition of convergence. 2. **Definition of convergence:** A sequence $$a_n$$ converges to a limit $$L$$ if for every $$\epsilon > 0$$, there exists an integer $$N$$ such that for all $$n > N$$, $$|a_n - L| < \epsilon$$. 3. **Apply the definition:** Here, $$a_n = \frac{2n}{2+n}$$ and $$L = 2$$. 4. **Calculate the difference:** $$ |a_n - L| = \left| \frac{2n}{2+n} - 2 \right| = \left| \frac{2n - 2(2+n)}{2+n} \right| = \left| \frac{2n - 4 - 2n}{2+n} \right| = \left| \frac{-4}{2+n} \right| = \frac{4}{2+n} $$ 5. **Find $$N$$ for given $$\epsilon$$:** We want $$ \frac{4}{2+n} < \epsilon \implies 2 + n > \frac{4}{\epsilon} \implies n > \frac{4}{\epsilon} - 2 $$ 6. **Conclusion:** For any $$\epsilon > 0$$, choose $$N = \left\lceil \frac{4}{\epsilon} - 2 \right\rceil$$. Then for all $$n > N$$, $$ |a_n - 2| < \epsilon $$ which proves $$\lim_{n \to \infty} \frac{2n}{2+n} = 2$$ by the definition of convergence.