Let's do some calculus! (2)

Since $0 \le \sin^2 x \le 1, \ \forall x \in \mathbb R$

$\implies 0 \le \dfrac {n^k\sin^2 (n!)}{n+2} \le \dfrac {n^k}{n+2}$ .

We note that $\displaystyle \lim_{n \to \infty} \dfrac {n^k}{n+2} = 0, \ k \in (0, 1)$

$\implies 0 \le \displaystyle \lim_{n \to \infty} \dfrac {n^k\sin^2 (n!)}{n+2} \le 0$

And by squeeze theorem , we have $\displaystyle \lim_{n \to \infty} \dfrac {n^k\sin^2 (n!)}{n+2} = \boxed{0}$ .