The value of mathop {text{Limit}}limits_{x to | vedclass

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Question

(A) Let the given limit be $L = \mathop {	ext{Limit}}\limits_{x 	o \infty } \frac{{\cot ^{ - 1}}\left( \frac{\log _a x}{x^a} ight)}{{\sec ^{ - 1}}

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Let the given limit be $L = \mathop {	ext{Limit}}\limits_{x 	o \infty } \frac{{\cot ^{ - 1}}\left( \frac{\log _a x}{x^a} ight)}{{\sec ^{ - 1}}\left( \frac{a^x}{\log _a x} ight)}$.As $x 	o \infty$,$\frac{\log _a x}{x^a} 	o 0$ because exponential growth dominates logarithmic growth for $a > 1$.Thus,the numerator approaches $\cot ^{ - 1}(0) = \pi / 2$.As $x 	o \infty$,$\frac{a^x}{\log _a x} 	o \infty$ because exponential growth dominates logarithmic growth.Thus,the denominator approaches $\sec ^{ - 1}(\infty) = \pi / 2$.Therefore,$L = \frac{\pi / 2}{\pi / 2} = 1$.

The value of $\mathop {\text{Limit}}\limits_{x \to \infty } \frac{{\cot ^{ - 1}}\left( {{x^{ - a}}\log _a x} \right)}{{\sec ^{ - 1}}\left( {{a^x}\log _x a} \right)}$ for $a > 1$ is equal to

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