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NTA Abhyas JEE Main2020MathematicsLimitsPractice

The value of lim x → 0 ⁡ log ⁡ 1 + 2 x 5 x + lim x → 2 ⁡ x 4 - 2 4 x - 2 is equal to

Correct answer

32.4

Step-by-step solution

lim x → 0 ⁡ log ⁡ 1 + 2 x 5 x + lim x → 0 ⁡ x 4 - 2 4 x - 2 = lim x → 0 log 1 + 2 x 2 x ⋅ 2 5 + lim x → 2 4 ⋅ x 3 = 2 5 + 32 = 32.4

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