Question

$\underset{x\to \infty }{lim}\frac{{\cot }^{-1}(\sqrt{x+1}-\sqrt{x})}{{\sec }^{-1}\left\{{\left(\frac{2x+1}{x-1}\right)}^x\right\}}$ is equal to

• A. $1$
• B. $0$
• C. $\frac{\pi }{2}$
• D. non existent

Answer 1

The correct option is A $1$

$\underset{x\to \infty }{lim}\frac{{\cot }^{-1}(\sqrt{x+1}-\sqrt{x})}{{\sec }^{-1}\left\{{\left(\frac{2x+1}{x-1}\right)}^x\right\}}$
$=\underset{x\to \infty }{lim}\frac{{\cot }^{-1}\left[\right(\sqrt{x+1}-\sqrt{x}).\frac{\sqrt{x+1}+\sqrt{x}}{\sqrt{x+1}+\sqrt{x}}]}{{\sec }^{-1}\left\{{\left(\frac{2x+1}{x-1}\right)}^x\right\}}$
$=\underset{x\to \infty }{lim}\frac{{\cot }^{-1}\left[\frac{1}{\sqrt{x+1}+\sqrt{x}}\right]}{{\sec }^{-1}\left\{{\left(\frac{2x+1}{x-1}\right)}^x\right\}}$
$=\underset{x\to \infty }{lim}\frac{{\tan }^{-1}[\sqrt{x+1}+\sqrt{x}]}{{\cos }^{-1}\left\{{\left(\frac{x-1}{2x+1}\right)}^x\right\}}=\frac{{\tan }^{-1}\left(\infty \right)}{{\cos }^{-1}\left(0\right)}=\frac{\pi /2}{\pi /2}=1$