NTA Abhyas JEE Main2020MathematicsDefinite IntegrationPractice
The limit L = lim n → ∞ ∑ r = 4 n - 4 n n 2 + r 2 satisfies the relation
Options
- Ae L > e
- Be L > 3
- Ce tan L < 2 e
- Dπ L < 1
Correct answer
C. e tan L < 2 e
Step-by-step solution
Given limits is L = lim n → ∞ ⁡ ∑ r = 4 n - 4 1 1 + r n 2 ⋅ 1 n = ∫ 0 1 d x 1 + x 2 = tan - 1 ⁡ x 0 1 = π 4 Hence, e tan π 4 = e < 2 e