Limit _ x , ^ - 1 ( x + 1 , , - , x ) ^ - 1 ( 2x + 1 x - 1 ) ^x i… - Vidyadip

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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