$\lim\limits_{x\to 0}\large\frac{\cos(\sin x)-1}{x^2}\qquad\big(\large\frac{0}{0}\big)$ form
$\Rightarrow \lim\limits_{x\to 0}\large\frac{-\sin(\sin x)\cos x}{2x}$
$\Rightarrow \large\frac{1}{2}$$\lim\limits_{x\to 0}\large\frac{\sin(\sin x)}{\sin x}.\big(\large\frac{\sin x}{x}\big)$$.\cos x$
$\Rightarrow \large\frac{-1}{2}$
Hence (d) is the correct answer.