1. **Problem statement:** Show that $$\lim_{n \to \infty} \frac{2n+1}{n^2} = 0$$ using the definition of limit. 2. **Recall the definition of limit for sequences:** For every $$\varepsilon > 0$$, there exists an $$N$$ such that for all $$n > N$$, $$\left| \frac{2n+1}{n^2} - 0 \right| < \varepsilon$$. 3. **Start with the inequality:** $$\left| \frac{2n+1}{n^2} - 0 \right| = \frac{2n+1}{n^2} < \varepsilon$$ 4. **Multiply both sides by $$n^2$$ (positive, so inequality direction preserved):** $$2n + 1 < \varepsilon n^2$$ 5. **Rewrite as:** $$\varepsilon n^2 - 2n - 1 > 0$$ 6. **Divide entire inequality by $$\varepsilon$$ (positive):** $$n^2 - \frac{2}{\varepsilon} n - \frac{1}{\varepsilon} > 0$$ 7. **Complete the square:** $$n^2 - \frac{2}{\varepsilon} n = \left(n - \frac{1}{\varepsilon}\right)^2 - \left(\frac{1}{\varepsilon}\right)^2$$ So the inequality becomes: $$\left(n - \frac{1}{\varepsilon}\right)^2 - \left(\frac{1}{\varepsilon}\right)^2 - \frac{1}{\varepsilon} > 0$$ 8. **Simplify:** $$\left(n - \frac{1}{\varepsilon}\right)^2 > \left(\frac{1}{\varepsilon}\right)^2 + \frac{1}{\varepsilon} = \frac{1 + \varepsilon}{\varepsilon^2}$$ 9. **Take square roots:** $$\left| n - \frac{1}{\varepsilon} \right| > \sqrt{\frac{1 + \varepsilon}{\varepsilon^2}} = \frac{\sqrt{1 + \varepsilon}}{\varepsilon}$$ 10. **Choose $$N$$ such that:** $$N > \frac{1}{\varepsilon} + \frac{\sqrt{1 + \varepsilon}}{\varepsilon}$$ Then for all $$n > N$$, the inequality holds, so $$\left| \frac{2n+1}{n^2} - 0 \right| < \varepsilon$$. **Final answer:** $$\lim_{n \to \infty} \frac{2n+1}{n^2} = 0$$ This completes the convergence proof using the epsilon definition of limit.