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

Solution:

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

$=\underset{x\to \mathrm{\infty }}{lim} \frac{{\cot }^{-1}\left(\frac{1}{\sqrt{x+1}+\sqrt{x}}\right)}{{\sec }^{-1}\left\{{\left(2+\frac{3}{x-1}\right)}^x\right\}}=\frac{{\cot }^{-1}\left(0\right)}{{\sec }^{-1}\left(\mathrm{\infty }\right)}=\frac{\pi /2}{\pi /2}=1$