MCQ
$\mathop {Limit}\limits_{x \to \infty } \,\frac{{{{\cot }^{ - 1}}\left( {\sqrt {x + 1} \,\, - \,\sqrt x } \right)}}{{{{\sec }^{ - 1}}\left\{ {{{\left( {\frac{{2x + 1}}{{x - 1}}} \right)}^x}} \right\}}}$ is equal to
  • $1$
  • B
    $0$
  • C
    $\pi /2$
  • D
    non existent

Answer

Correct option: A.
$1$
a
$\mathop {Limit}\limits_{x \to \infty } \,\sqrt {x + 1} \, - \sqrt x \,$ $= 0$  $\Rightarrow cot^{-1}(0) = \pi /2$
$\mathop {Limit}\limits_{x \to \infty } \,{\left( {\frac{{2x + 1}}{{x - 1}}} \right)^x}\,\, = \,\,\infty \,$ $\Rightarrow\,\,\, sec^{-1} (\infty ) = \pi /2$                     

$\therefore\,\, l = 1$ Ans 

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