mathop {Limit}limits_{x to infty } ,frac{{{{c | vedclass

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(A) First,evaluate the numerator: $\mathop {Limit}\limits_{x 	o \infty } (\sqrt {x + 1} - \sqrt x) = \mathop {Limit}\limits_{x 	o \infty } \frac{(x+

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(A) First,evaluate the numerator: $\mathop {Limit}\limits_{x 	o \infty } (\sqrt {x + 1} - \sqrt x) = \mathop {Limit}\limits_{x 	o \infty } \frac{(x+1)-x}{\sqrt{x+1}+\sqrt{x}} = \mathop {Limit}\limits_{x 	o \infty } \frac{1}{\sqrt{x+1}+\sqrt{x}} = 0$. Thus,$\cot^{-1}(0) = \pi / 2$. Next,evaluate the denominator: $\mathop {Limit}\limits_{x 	o \infty } (\frac{2x+1}{x-1})^x = \mathop {Limit}\limits_{x 	o \infty } (2 + \frac{3}{x-1})^x = \infty$. Thus,$\sec^{-1}(\infty) = \pi / 2$. Therefore,the limit is $\frac{\pi / 2}{\pi / 2} = 1$.

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

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