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JEE Main 2013

JEE Main 2013 Mathematics question on Limits Continuity and Differentiability

lim⁡x→0(1−cos⁡2x)(3+cos⁡x)xtan⁡4x\lim_{x \to 0} \dfrac{(1 - \cos 2x)(3 + \cos x)}{x \tan 4x} is equal to
  1. A.
    4
  2. B.
    3
  3. C.
    2
  4. D.
    12\dfrac{1}{2}
Show answer and solution

Correct option: C

Solution

Use 1 - cos2x = 2sin²x and tan4x ≈ 4x as x→0.

The limit becomes: (2sin²x · (3 + cos x)) / (x · 4x) = (2sin²x / x²) · (3 + cos x) / 4.

As x→0, sin²x/x² → 1 and cos x → 1, so the limit = 2 · (3+1)/4 = 2 · 1 = 2.

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