1. The problem is to calculate the limit $$\lim_{n \to \infty} (\sqrt{2n + 1} - \sqrt{n + 2})$$. 2. To solve limits involving differences of square roots, we use the conjugate to simplify: $$\sqrt{a} - \sqrt{b} = \frac{(\sqrt{a} - \sqrt{b})(\sqrt{a} + \sqrt{b})}{\sqrt{a} + \sqrt{b}} = \frac{a - b}{\sqrt{a} + \sqrt{b}}$$ 3. Apply this to our expression: $$\lim_{n \to \infty} \frac{(2n + 1) - (n + 2)}{\sqrt{2n + 1} + \sqrt{n + 2}} = \lim_{n \to \infty} \frac{n - 1}{\sqrt{2n + 1} + \sqrt{n + 2}}$$ 4. Factor $n$ in numerator and denominator to analyze the limit: $$\lim_{n \to \infty} \frac{n(1 - \frac{1}{n})}{\sqrt{n}\sqrt{2 + \frac{1}{n}} + \sqrt{n}\sqrt{1 + \frac{2}{n}}} = \lim_{n \to \infty} \frac{n(1 - \frac{1}{n})}{\sqrt{n}(\sqrt{2 + \frac{1}{n}} + \sqrt{1 + \frac{2}{n}})}$$ 5. Simplify by canceling $\sqrt{n}$: $$\lim_{n \to \infty} \frac{\cancel{n}(1 - \frac{1}{n})}{\cancel{\sqrt{n}}(\sqrt{2 + \frac{1}{n}} + \sqrt{1 + \frac{2}{n}})} = \lim_{n \to \infty} \frac{\cancel{\sqrt{n}}\sqrt{n}(1 - \frac{1}{n})}{\cancel{\sqrt{n}}(\sqrt{2 + \frac{1}{n}} + \sqrt{1 + \frac{2}{n}})} = \lim_{n \to \infty} \frac{\sqrt{n}(1 - \frac{1}{n})}{\sqrt{2 + \frac{1}{n}} + \sqrt{1 + \frac{2}{n}}}$$ 6. As $n \to \infty$, $1 - \frac{1}{n} \to 1$, $\sqrt{2 + \frac{1}{n}} \to \sqrt{2}$, and $\sqrt{1 + \frac{2}{n}} \to 1$. 7. So the limit behaves like: $$\lim_{n \to \infty} \frac{\sqrt{n} \times 1}{\sqrt{2} + 1} = \infty$$ 8. Therefore, the limit diverges to infinity. **Final answer:** $$\lim_{n \to \infty} (\sqrt{2n + 1} - \sqrt{n + 2}) = \infty$$