Question

Among $\left(i\right)\underset{x\to \infty }{lim}se{c}^{-1}\left(\frac{x}{\sin x}\right)$ and $\left(ii\right)\underset{x\to \infty }{lim}se{c}^{-1}\left(\frac{\sin x}{x}\right)$

• A. $\left(i\right)$ exists, $\left(ii\right)$ does not exist
• B. $\left(i\right)$ does not exist, $\left(ii\right)$ exists
• C. Both $\left(i\right)$ and $\left(ii\right)$ exist
• D. Neither $\left(i\right)$ nor $\left(ii\right)$ exists

Answer 1

The correct option is A $\left(i\right)$ exists, $\left(ii\right)$ does not exist

$i.\underset{x\to \infty }{lim}{\sec }^{-1}\left(\frac{x}{\sin x}\right)={\sec }^{-1}\left(\frac{\infty }{\sin \infty }\right)$
$={\sec }^{-1}\left(\frac{\infty }{\text{any value between -1 to 1}}\right)$
$=se{c}^{-1}(\pm \infty )=\frac{\pi }{2}$
$ii.\underset{x\to \infty }{lim}se{c}^{-1}\left(\frac{\sin x}{x}\right)=se{c}^{-1}\left(\frac{\sin \infty }{\infty }\right)$
$={\sec }^{-1}\left(\frac{\text{any value between -1 to 1}}{\infty }\right)$
$=se{c}^{-1}0=\text{not defined}$
$\text{Hence, (i) exists but (ii) does not exist.}$