MCQ
$\mathop {\lim }\limits_{x \to \infty } \frac{{{{\cot }^{ - 1}}\left( {\sqrt {x + 1} - \sqrt x } \right)}}{{{{\sec }^{ - 1}}\left\{ {{{\left( {\frac{{2x + 1}}{{x - 1}}} \right)}^x}} \right\}}}$ is equal to-
- ✓$1$
- B$0$
- C$\pi /2$
- Dnon existent
$=\lim _{x \rightarrow \infty} \frac{\cot ^{-1}\left[(\sqrt{x+1}-\sqrt{x}) \cdot \frac{\sqrt{x+1}+\sqrt{x}}{\sqrt{x+1}+\sqrt{x}}\right]}{\sec ^{-1}\left\{\left(\frac{2 x+1}{x-1}\right)^{x}\right\}}$
$ =\lim _{x \rightarrow \infty} \frac{\cot ^{-1}\left[\frac{1}{\sqrt{x+1}+\sqrt{x}}\right]}{\sec ^{-1}\left\{\left(\frac{2 x+1}{x-1}\right)^{x}\right\}}$
$= \lim _{x \rightarrow \infty} \frac{\tan ^{-1}[\sqrt{x+1}+\sqrt{x}]}{\cos ^{-1}\left\{\left(\frac{x-1}{2 x+1}\right)^{x}\right\}}=\frac{\tan ^{-1}(\infty)}{\cos ^{-1}(0)}=\frac{\pi / 2}{\pi / 2}=1 $
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