Limits
Trigonometric Limits
JEE Main 2019
Grade 11

Question:

$\lim_{x \to 0} \frac{x \cot 4x}{\sin^2 x \cdot \cot^2 2x}$
(1) $0$
(2) $1$
(3) $4$
(4) $\frac{1}{4}$

Step-by-Step Solution

Key Concept: Write $\cot = \cos / \sin$ throughout. The expression becomes $\frac{x \cos 4x \sin^2 2x}{\sin^2 x \cdot \sin 4x \cos^2 2x}$. Near $x = 0$: $\sin \theta \approx \theta$ and $\cos \theta \approx 1$, giving $\frac{x \cdot (2x)^2}{x^2 \cdot 4x} = \frac{4x^3}{4x^3} = 1$.
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Correct Answer: (2)

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