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JEE Main 2014

JEE Main 2014 Mathematics question on Limits Continuity and Differentiability

lim⁡x→0sin⁡(πcos⁡2x)x2\lim_{x \to 0} \frac{\sin(\pi \cos^2 x)}{x^2} is equal to
  1. A.
    π2\frac{\pi}{2}
  2. B.
    11
  3. C.
    −π-\pi
  4. D.
    π\pi
Show answer and solution

Correct option: D

Solution

As x→0, cos²x → 1, so π cos²x → π. Write π cos²x = π - π sin²x, so sin(π cos²x) = sin(π - π sin²x) = sin(π sin²x).

Now the limit becomes lim(x→0) sin(π sin²x)/x². Since sin²x ≈ x² near 0, this equals lim(x→0) sin(π sin²x)/(π sin²x) · π sin²x/x² = 1 · π · 1 = π.

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