$\lim_{n \to \infty} n \sin\left(\frac{2\pi}{\sqrt{n^2 + 1}}\right)$
Step-by-Step Solution
Key Concept: As $n \to \infty$, $\frac{2\pi}{\sqrt{n^2 + 1}} \approx \frac{2\pi}{n} \to 0$. So $n \sin\left(\frac{2\pi}{\sqrt{n^2 + 1}}\right) \approx n \cdot \frac{2\pi}{\sqrt{n^2 + 1}} \to \frac{2\pi n}{n} = 2\pi$.
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Correct Answer: (3)