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NTA Abhyas JEE Main2020MathematicsLimitsPractice

The value of lim n → ∞ ⁡ cos ⁡ x 2 cos ⁡ x 4 cos ⁡ x 8 … … . . . cos ⁡ x 2 n + 1 is equal to

Options

  1. Ax sin ⁡ x
  2. Bsin ⁡ x x
  3. C0
  4. DNone of these

Correct answer

B. sin ⁡ x x

Step-by-step solution

Let, L = lim n → ∞ ⁡ cos ⁡ x 2 cos ⁡ x 2 2 cos ⁡ x 2 3 … . . cos ⁡ x 2 n + 1 = lim n → ∞ ⁡ 1 2 sin ⁡ x 2 n + 1 cos ⁡ x 2 … … cos ⁡ x 2 n 2 sin ⁡ x 2 n + 1 cos ⁡ x 2 n + 1 = l i m n → ∞ 1 2 2 sin ⁡ x 2 n + 1 cos ⁡ x 2 … 2 cos ⁡ x 2 n sin ⁡ x 2 n = lim n → ∞ ⁡ 1 2 n + 1 sin ⁡ x 2 n + 1 2 cos ⁡ x 2 sin ⁡ x 2 = lim n → ∞ &#8

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