NTA Abhyas JEE Main2020MathematicsLimitsPractice

The value of l i m x → 0 l n 2 - cos ⁡ 15 x l n 2 sin ⁡ 3 x + 1 is equal to

Correct answer

12.5

Step-by-step solution

l i m x → 0 l n 2 - cos ⁡ 15 x l n 2 sin ⁡ 3 x + 1 = l i m x → 0 l n 1 + 1 - cos ⁡ 15 x l n 2 1 + sin ⁡ 3 x = l i m x → 0 1 - cos ⁡ 15 x sin ⁡ 3 x 2 = l i m x → 0 15 x 2 2 3 x 2 = 1 225 2 × 9 = 12 . 5

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