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JEE Main 2019 · 11 Jan · Shift II

JEE Main 2019 Mathematics question on Limits Continuity and Differentiability

lim⁡x→0xcot⁡(4x)sin⁡2xcot⁡2(2x)\lim_{x \to 0} \dfrac{x\cot(4x)}{\sin^2 x \cot^2(2x)} is equal to
  1. A.
    0
  2. B.
    1
  3. C.
    4
  4. D.
    2
Show answer and solution

Correct option: B

Solution

As x→0, use the small angle approximations: cot(4x) ≈ 1/(4x), sin x ≈ x, cot(2x) ≈ 1/(2x).

Substituting: numerator = x · 1/(4x) = 1/4, denominator = x² · 1/(4x²) = 1/4.

Limit = (1/4)/(1/4) = 1.

[JEE Main 2019 · 11 Jan · Shift II]

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