Question

$\underset{x\to \infty }{lim}{\sec }^{-1}\left(\frac{x}{x+1}\right)$ is equal to

• A. $0$
• B. $\pi$
• C. $\frac{\pi }{2}$
• D. does not exist

Answer 1

The correct option is A $0$

Given that

${lim}_{x\to \infty }{\sec }^{-1}\left(\frac{x}{x+1}\right)$

$={lim}_{x\to \infty }{\cos }^{-1}\left(\frac{x+1}{x}\right)$ Since ${\cos }^{-1}x={\sec }^{-1}\frac{1}{x}$

$={lim}_{x\to \infty }{\cos }^{-1}(\frac{x}{x}+\frac{1}{x})$

$={lim}_{x\to \infty }{\cos }^{-1}(1+\frac{1}{x})$

Put the limit and we get

$={\cos }^{-1}(1+\frac{1}{\infty })$

$={\cos }^{-1}(1+0)$

$={\cos }^{-1}1$

$={\cos }^{-1}\cos 0$

$=0$

Hence, option $A$ is correct answer