1. The problem asks: Determine whether the sequence \(\left\{ \frac{\sqrt{n + 47} - \sqrt{n}}{1} \right\}_{n=0}^\infty\) converges or diverges and compute its limit if it converges. 2. We start by simplifying the expression \(a_n = \sqrt{n+47} - \sqrt{n}\). 3. Use the conjugate to simplify: $$a_n = \sqrt{n+47} - \sqrt{n} = \frac{(\sqrt{n+47} - \sqrt{n})(\sqrt{n+47} + \sqrt{n})}{\sqrt{n+47} + \sqrt{n}} = \frac{(n+47) - n}{\sqrt{n+47} + \sqrt{n}} = \frac{47}{\sqrt{n+47} + \sqrt{n}}$$ 4. Now consider the limit as \(n \to \infty\): $$\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{47}{\sqrt{n+47} + \sqrt{n}}$$ 5. For large \(n\), \(\sqrt{n+47} \approx \sqrt{n}\), so denominator \(\approx \sqrt{n} + \sqrt{n} = 2\sqrt{n}\). 6. Therefore, $$\lim_{n \to \infty} a_n \approx \lim_{n \to \infty} \frac{47}{2\sqrt{n}} = 0$$ 7. Since \(a_n > 0\) for all \(n\) and the terms tend toward zero, the sequence converges and its limit is 0.