Evaluate $\lim_{n \to \infty} \frac{\sin(\sqrt{n})-\sin\sqrt{n-1}}{n^1}$
Step-by-Step Solution
Key Concept: Sum-to-product formula for sine differences combined with asymptotic analysis of $\sqrt{n} - \sqrt{n-1}$.
Using the identity $\sin A - \sin B = 2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$, we have $\sin(\sqrt{n}) - \sin(\sqrt{n-1}) = 2\cos\left(\frac{\sqrt{n}+\sqrt{n-1}}{2}\right)\sin\left(\frac{\sqrt{n}-\sqrt{n-1}}{2}\right)$. Since $\sqrt{n} - \sqrt{n-1} = \frac{1}{\sqrt{n}+\sqrt{n-1}} \approx \frac{1}{2\sqrt{n}}$, the sine term behaves like $\frac{1}{4\sqrt{n}}$. The cosine is bounded by 1, so the numerator is $O(n^{-1/2})$. Dividing by $n$ gives $O(n^{-3/2}) \to 0$.
Correct Answer: 0