limn→∞sinπ√(n^2+1) is equal to

Correct answer: (B) 0. limn→∞sinnπ1+1n2^(1/2) =limn→∞sinnπ1+1/(2n^2)-1/(8n^4)+… =limn→∞sinnπ+(π)/2n-(π)/(8n^3)+… =limn→∞-1^nsinπ1/2n-1/(8n^3)+…=0

Q59 / 125 · BITSAT 2017 · Mathematics

$\lim_{n \to \infty} \sin [π \sqrt{n^{2} + 1}]$ is equal to

$\lim_{n \to \infty} \sin \{n π {(1 + \frac{1}{n^{2}})}^{1 / 2}\}$

$= \lim_{n \to \infty} \sin \{n π (1 + \frac{1}{2 n^{2}} - \frac{1}{8 n^{4}} + \dots)\}$

$= \lim_{n \to \infty} \sin \{n π + \frac{π}{2 n} - \frac{π}{8 n^{3}} + \dots\}$

$= \lim_{n \to \infty} {(- 1)}^{n} \sin π (\frac{1}{2 n} - \frac{1}{8 n^{3}} + \dots) = 0$

Hence, the correct answer is option: 'B'.