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JEE Main201912 Apr 2019Evening ShiftMathematicsLimitsActual

l i m x → 0 x + 2 s i n x x 2 + 2 s i n x + 1 - s i n 2 x - x + 1 is

Options

  1. A3
  2. B1
  3. C2
  4. D6

Correct answer

C. 2

Step-by-step solution

lim x → 0 x + 2 sin x x 2 + 2 sin x + 1 − sin x − x + 1 0 0 form By using L'Hospital Rule, = lim x → 0 1 + 2 cos x 1 2 x + 2 cos x 2 x 2 + 2 sin x + 1 − 1 sin 2 x − 1 2 sin 2 x − x + 1 = 1 + 2 1 + 1 2 = 3 3 2 = 2 .

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