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JEE Main20238 Apr 2023Morning ShiftMathematicsLimitsActual

lim x → 0 1 - cos 2 ( 3 x ) cos 3 ( 4 x ) sin 3 ( 4 x ) log e 2 x + 1 5 is equal to

Options

  1. A15
  2. B9
  3. C18
  4. D24

Correct answer

C. 18

Step-by-step solution

Given, lim x → 0 1 - cos 2 ( 3 x ) cos 3 ( 4 x ) sin 3 ( 4 x ) log e 2 x + 1 5 Now we now that, lim x → 0 sin x x = 1 ,   lim x → 0 1 - cos x x 2 = 1 2   &   lim x → 0 log 1 + x x = 1 Now using the above formula we get, lim x → 0 1 - cos 2 3 x cos 3 4 x sin 3 4 x log e 2 x + 1 5 = lim x → 0 1 - cos 3 x 1 + cos 3 x 9 x 2 cos 3 4 x 9 x 2 sin 4 x 3 64 x 3 2 x 5 64 x 3 log e ( 2 x + 1 ) 5 2 x 5 = lim x → 0 1 - cos 3 x 9 x 2 1 + cos 3 x cos 3 4 x sin 4 x 4 x 3

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