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)
Answer & explanation
Correct answer: option 1
$\frac{x}{\sin x}$ 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.