Among $\lim\limits_{x \rightarrow 0} \sec ^{-1}\left(\frac{x}{\sin x}\right)$    .......(1)

and $\lim\limits_{x \rightarrow 0} \sec ^{-1}\left(\frac{\sin x}{x}\right)$    .......(2)

(1) exists, (2) does not exist

$\frac{x}{\sin x}$ is more than 1 in the neighbourhood of '0'. Hence $\lim\limits_{x \rightarrow 0} \sec ^{-1}\left(\frac{x}{\sin x}\right)$ exists while $\frac{\sin x}{x}$ is less than 1 in the neighbourhood of '0' . Hence $\lim\limits_{x \rightarrow 0} \sec ^{-1}\left(\frac{\sin x}{x}\right)$ does not exist.

Hence (1) is the correct answer.