NTA Abhyas JEE Main2020MathematicsLimitsPractice
The value of l i m x → 0 l n 10 - 9 cos ⁡ 2 x l n 2 sin ⁡ 3 x + 1 is equal to
Correct answer
2
Step-by-step solution
l i m x → 0 l n 10 - 9 cos ⁡ 2 x l n 2 sin ⁡ 3 x + 1 = l i m x → 0 l n 1 + 9 - 9 cos ⁡ 2 x l n 2 1 + sin ⁡ 3 x = l i m x → 0 9 1 - cos ⁡ 2 x sin ⁡ 3 x 2 = l i m x → 0 9 2 x 2 3 x 2 = 2