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The limit L = lim n → ∞ ⁡ ∑ r = 4 n - 4 n n 2 + r 2 satisfies the relation

Options

  1. Ae L > e
  2. Be L > 3
  3. Ce tan ⁡ L < 2 e
  4. Dπ L < 1

Correct answer

C. e tan ⁡ L < 2 e

Step-by-step solution

Given limits is L = lim n &#8594; &#8734; &#8289; &#8721; r = 4 n - 4 1 1 + r n 2 &#8901; 1 n = &#8747; 0 1 d x 1 + x 2 = tan - 1 &#8289; x 0 1 = &#960; 4 Hence, e tan &#960; 4 = e &#60; 2 e

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