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Limits — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
The value of lim⁡x→∞cot⁡−1(x−alog⁡ax)sec⁡−1(axlog⁡xa)\displaystyle\lim_{x\to\infty}\dfrac{\cot^{-1}(x^{-a}\log_{a}x)}{\sec^{-1}(a^{x}\log_{x}a)} (a>1a>1) is
A11correct
B00
Cπ2\dfrac{\pi}{2}
DDoes not exist
Solution
Step 1: As x→∞x\to\infty with a>1a>1: Numerator: x−alog⁡ax→0x^{-a}\log_{a}x\to 0 (since x−a→0x^{-a}\to 0 faster than log⁡ax→∞\log_{a}x\to\infty). So cot⁡−1(x−alog⁡ax)→cot⁡−1(0)=π2\cot^{-1}(x^{-a}\log_{a}x)\to\cot^{-1}(0)=\dfrac{\pi}{2}. Step 2: Denominator: axlog⁡xa=ax⋅1log⁡ax=axlog⁡ax→∞a^{x}\log_{x}a=a^{x}\cdot\dfrac{1}{\log_{a}x}=\dfrac{a^{x}}{\log_{a}x}\to\infty. So sec⁡−1(axlog⁡xa)→sec⁡−1(∞)=π2\sec^{-1}(a^{x}\log_{x}a)\to\sec^{-1}(\infty)=\dfrac{\pi}{2}. Step 3: Ratio:
π/2π/2=1\dfrac{\pi/2}{\pi/2}=1
 Correct answer: (1) 11
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