Explanation

[Solution by Dr. Abubacker] The bracket tends to $\sqrt5-\sqrt5=0$ and $\operatorname{cosec}x\to\infty$, so apply L'Hôpital to $g(x)/\sin x$: $g'(0)=\frac{-4\cos x\sin x-3\sin x}{2\sqrt{2\cos^2x+3\cos x}}\Big|_0-\frac{-2\cos x\sin x+\cos x}{2\sqrt{\cos^2x+\sin x+4}}\Big|_0=0-\frac1{2\sqrt5}$. The limit is $-\frac1{2\sqrt5}$. Answer (4).