1. The problem asks us to prove the limit $$\lim_{n \to \infty} \left(2 - \frac{1}{n}\right) = 2.$$\n\n2. The formula for the limit of a sequence $$a_n$$ as $$n$$ approaches infinity is: $$\lim_{n \to \infty} a_n = L$$ if for every $$\epsilon > 0$$, there exists an $$N$$ such that for all $$n > N$$, $$|a_n - L| < \epsilon$$.\n\n3. Here, our sequence is $$a_n = 2 - \frac{1}{n}$$ and we want to show it approaches $$L = 2$$.\n\n4. Calculate the absolute difference: $$|a_n - 2| = \left|2 - \frac{1}{n} - 2\right| = \left| - \frac{1}{n} \right| = \frac{1}{n}.$$\n\n5. Given any $$\epsilon > 0$$, choose $$N > \frac{1}{\epsilon}$$. Then for all $$n > N$$, $$\frac{1}{n} < \frac{1}{N} < \epsilon$$.\n\n6. This means $$|a_n - 2| < \epsilon$$ for all $$n > N$$, proving that $$\lim_{n \to \infty} \left(2 - \frac{1}{n}\right) = 2$$.\n\n7. In simple terms, as $$n$$ gets larger and larger, $$\frac{1}{n}$$ gets closer and closer to zero, so $$2 - \frac{1}{n}$$ gets closer and closer to 2.