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NTA Abhyas JEE Main2020MathematicsLimitsPractice

The value of lim x → 0 1 + sin x - cos x + ln 1 - x x · tan 2 x is

Options

  1. A- 1 2
  2. B- 1 3
  3. C1 2
  4. D1 4

Correct answer

A. - 1 2

Step-by-step solution

lim x → 0 1 + sin x - cos x + ℓ n 1 - x x · tan 2 x = lim x → 0 1 + sin x - cos x + ℓ n 1 - x x 3 0 0 [ ∵   lim x → 0 x tan   x = 1 ] = lim x → 0 cos x + sin x - 1 1 - x 3 x 2                   0 0 [using L'hospital rule] = lim x → 0 - sin x + cos x - 1 ( 1 - x ) 2 6 x                   0 0 [Using L'hospital rule again] = lim x → 0 - cos x - sin x -

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