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NTA Abhyas JEE Main2020MathematicsLimitsPractice

Let lim x → 0 ⁡ sin ⁡ 2 x tan ⁡ x k = L 1 and lim x → 0 ⁡ e 2 x - 1 x = L 2 , and the value of L 1 L 2 is 8, then k is

Options

  1. A4
  2. B8
  3. C6
  4. D2

Correct answer

D. 2

Step-by-step solution

L 1 = lim x → 0 ⁡ sin ⁡ 2 x 2 x ⋅ 2 x tan ⁡ x k x k ⋅ x k = 2 k L 2 = lim x → 0 ⁡ e 2 x - 1 x = lim x → 0 ⁡ e 2 x - 1 2 x ⋅ 2 = 2 ⇒ L 1 L 2 = 8 ⇒ 4 k = 8 ⇒ k = 2

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