$\lim\limits_{n \rightarrow \infty} \frac{n^k \sin ^2 n !}{n+2}$, 0 < k < 1, is equal to:

0

$\lim\limits_{n \rightarrow \infty} \frac{n^k \sin ^2 n !}{n+2}=\lim\limits_{n \rightarrow \infty} \frac{n^k \sin ^2 n !}{n(1+2 / n)}$

$=\lim\limits_{n \rightarrow \infty} \frac{\sin ^2(n !)}{n^{1-k}\left(1+\frac{2}{n}\right)}=\frac{\text { a finite quantity }}{\infty}$

[Since sin² n! always lies between 0 and 1. Also, since 1 - k > 0, hence n^{1 - k} → ∞

Hence (3) is the correct answer. as n→ ∞]