Question:
\(\lim\limits_{n \to \infty} (\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+\frac{n}{n^2+3^2}+…+\frac{1}{5n})=\)
\(\lim\limits_{n \to \infty} (\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+\frac{n}{n^2+3^2}+…+\frac{1}{5n})=\)
Updated On: Apr 26, 2024
- \(\frac{π}{4}\)
- \(tan^{-1}3\)
- \(tan^{-1}2\)
- \(\frac{π}{2}\)
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The Correct Option is C
Solution and Explanation
The correct answer is Option (C) : \(tan^{-1}2\)
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