JEE Main 202126 Aug, Morning shiftMathematicsEasyIntegral Calculus369× since 2002The value of limn→∞1n∑r=02n−1n2n2+4r2\mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 0}^{2n - 1} {{{{n^2}} \over {{n^2} + 4{r^2}}}}n→∞limn1r=0∑2n−1n2+4r2n2 is :AOption A. 12tan−1(2){1 \over 2}{\tan ^{ - 1}}(2)21tan−1(2)BOption B. 12tan−1(4){1 \over 2}{\tan ^{ - 1}}(4)21tan−1(4)COption C. tan−1(4){\tan ^{ - 1}}(4)tan−1(4)DOption D. 14tan−1(4){1 \over 4}{\tan ^{ - 1}}(4)41tan−1(4)CheckSkip · show solution
