limx→∞ cot−1⁡(x+1−x)sec−1⁡2x+1x−1x is equal to

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

1. A

1

2. B

0

3. C

$\frac{\pi }{2}$

4. D

non-existent

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### Solution:

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

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

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