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NTA Abhyas JEE Main2020MathematicsLimitsPractice

The value of lim x → 0 ⁡ 9 ln ⁡ 2 - cos ⁡ 25 x 5 ln 2 sin ⁡ 3 x + 1 ⁡ is equal to

Correct answer

62.5

Step-by-step solution

lim x → 0 ⁡ 9 ln 2 - cos ⁡ 25 x 5 ln 2 sin ⁡ 3 x + 1 = lim x → 0 ⁡ 9 ln 1 + 1 - cos ⁡ 25 x 5 ln 2 1 + sin ⁡ 3 x = lim x → 0 ⁡ 9 1 - cos ⁡ 25 x 5 sin ⁡ 3 x 2 ⇒ lim x → 0 ⁡ 9 25 x 2 10 3 x 2 = 625 10 = 62 . 5

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