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NTA Abhyas JEE Main2020MathematicsLimitsPractice

The value of lim x → ∞ ⁡ e 2 1 + 2 x x x is equal to

Options

  1. Ae 2
  2. Be - 1
  3. Ce 1 2
  4. De - 1 2

Correct answer

A. e 2

Step-by-step solution

Let, L = l i m x → ∞ e 2 1 + 2 x x x = l i m y → 0 + e 2 1 + 2 y 1 y 1 y   b y  p u t t i n g   1 x = y L = e lim y → 0 + 1 y ln e 2 1 + 2 y 1 y Now, we have, l i m y → 0 + 1 y l n e 2 1 + 2 y 1 y = l i m y → 0 + l n e 2 - 1 y l n 1 + 2 y y = l i m y → 0 + 2 y - l n 1 + 2 y y 2 = l i m y → 0 + 2 y - 2 y - 4 y 2 2 + 8 y 3 3 - 16 y 4 4 + … y 2 = l i m y → 0 + 4 2 + 8 y 3 + 16 y 2 4 - … = 2 Hence, L = e 2

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