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NTA Abhyas JEE Main2020MathematicsLimitsPractice

The value of l i m n → ∞ cos ⁡ x c o s x 2 c o s x 4 . . . c o s x 2 n is equal to

Options

  1. Ax sin ⁡ x
  2. Bsin ⁡ x x
  3. Csin ⁡ 2 x 2 x
  4. D2 x sin ⁡ 2 x

Correct answer

C. sin ⁡ 2 x 2 x

Step-by-step solution

Required limit = l i m n → ∞ c o s x 2 0 c o s x 2 1 c o s x 2 2 . . . c o s x 2 n = l i m n → ∞ 1 2 s i n x 2 n cos ⁡ x . . . c o s x 2 n - 1 2 s i n x 2 n c o s x 2 n = l i m n → ∞ 1 2 2 s i n x 2 n cos ⁡ x . . . 2 c o s x 2 n - 1 s i n x 2 n - 1 = l i m n → ∞ 1 2 n + 1 s i n x 2 n 2 cos ⁡ x sin ⁡ x = l i m n → ∞ sin ⁡ 2 x 2 n + 1 s i n x 2 n = sin ⁡ 2 x 2 x l i m n → ∞ x 2 n s i n x 2 n = sin ⁡ 2 x 2 x

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