mathop {lim }limits_{x to infty } frac{{{{cot | vedclass

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(A) Let $L = \lim _{x}$ ${\rightarrow \infty} \frac{\cot ^{-1}(\sqrt{x+1}-\sqrt{x})}{\sec ^{-1}\left\{\left(\frac{2 x+1}{x-1}\right)^{x}\right\}}$.Fir

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(A) Let $L = \lim _{x}$ ${ightarrow \infty} \frac{\cot ^{-1}(\sqrt{x+1}-\sqrt{x})}{\sec ^{-1}\left\{\left(\frac{2 x+1}{x-1}ight)^{x}ight\}}$.First,simplify the numerator:$\sqrt{x+1}-\sqrt{x} = \frac{(\sqrt{x+1}-\sqrt{x})(\sqrt{x+1}+\sqrt{x})}{\sqrt{x+1}+\sqrt{x}} = \frac{1}{\sqrt{x+1}+\sqrt{x}}$.As $x 	o \infty$,$\frac{1}{\sqrt{x+1}+\sqrt{x}} 	o 0$,so $\cot ^{-1}(\frac{1}{\sqrt{x+1}+\sqrt{x}}) 	o \cot ^{-1}(0) = \frac{\pi}{2}$.Next,examine the denominator:As $x 	o \infty$,$\frac{2x+1}{x-1} 	o 2$. Thus,$(\frac{2x+1}{x-1})^x 	o \infty$.Since $\sec ^{-1}(u)$ is defined for $|u| \ge 1$,and $\sec ^{-1}(\infty) = \frac{\pi}{2}$.Therefore,$L = \frac{\pi/2}{\pi/2} = 1$.

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

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