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NTA Abhyas JEE Main2020MathematicsLimitsPractice

The value of l i m x → 0 ln ⁡ 1 + 2 x + 4 x 2 + ln ⁡ 1 - 2 x + 4 x 2 sec ⁡ x - cos ⁡ x is equal to

Options

  1. A- 1
  2. B1
  3. C0
  4. D4

Correct answer

D. 4

Step-by-step solution

l i m x → 0 ln ⁡ 1 + 2 x + 4 x 2 + ln ⁡ 1 - 2 x + 4 x 2 sec ⁡ x - cos ⁡ x = l i m x → 0 ln ⁡ 1 + 4 x 2 2 - 4 x 2 1 - c o s 2 x / cos ⁡ x = l i m x → 0 ln ⁡ 1 + 4 x 2 + 16 x 4 sin ⁡ x tan ⁡ x = l i m x → 0 ln ⁡ 1 + 4 x 2 1 + 4 x 2 4 x 2 1 + 4 x 2 4 x 2 1 + 4 x 2 ⋅ 1 sin ⁡ x x ⋅ tan ⁡ x x ⋅ x 2 = 4 a s l i m x → 0 ln ⁡ 1 + x x = 1

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