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JEE Advanced · Mathematics · 16. Limits

The value of the limit limxπ242sin3x+sinx2sin2xsin3x2+cos5x2-2+2cos2x+cos3x2 is _________

  1. A 10
  2. B 15
  3. C 8
  4. D 3
Verified Solution

Answer & Solution

Correct Answer

(C) 8

Step-by-step Solution

Detailed explanation

limxπ282 sin2x·cosxcosx2-cos7x2+cos5x2-2·2cos2x+cos3x2 =limxπ282 sin2x·cosxcosx2-cos3x2+cos5x2-cos7x2-22cos2x
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