Question:
\( \lim_{x\to 0} \frac{x\cot(4x)}{\sin^2 x\,\cot^2(2x)} \) is equal to :
\( \lim_{x\to 0} \frac{x\cot(4x)}{\sin^2 x\,\cot^2(2x)} \) is equal to :
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Near 0, \( \cot kx \approx \frac{1}{kx} \) and \( \sin x \approx x \); substitute the leading terms in numerator and denominator.
Updated On: Nov 5, 2026
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The Correct Option is C
Solution and Explanation
As \( x \to 0 \): \( x\cot 4x = \frac{x\cos 4x}{\sin 4x} \to \frac{x}{4x} = \frac{1}{4} \).
\( \sin^2 x\cot^2 2x \approx x^2\cdot\frac{1}{(2x)^2} = \frac{1}{4} \) (more precisely \( \sin^2 x \cdot \frac{\cos^2 2x}{\sin^2 2x} \to \frac{x^2}{4x^2} \)).
So the limit \( = \frac{1/4}{1/4} = 1 \).
Answer: (3) 1
\( \sin^2 x\cot^2 2x \approx x^2\cdot\frac{1}{(2x)^2} = \frac{1}{4} \) (more precisely \( \sin^2 x \cdot \frac{\cos^2 2x}{\sin^2 2x} \to \frac{x^2}{4x^2} \)).
So the limit \( = \frac{1/4}{1/4} = 1 \).
Answer: (3) 1
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