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Q. Among (i) $\displaystyle\lim _{x \rightarrow \infty} \sec ^{-1}\left(\frac{x}{\sin x}\right)$ and (ii) $\displaystyle\lim _{x \rightarrow \infty} \sec ^{-1}\left(\frac{\sin x}{x}\right)$,

Limits and Derivatives

Solution:

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