JEE Main 202228 Jun, Evening shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002The value of limn→∞6tan{∑r=1ntan−1(1r2+3r+3)}\mathop {\lim }\limits_{n \to \infty } 6\tan \left\{ {\sum\limits_{r = 1}^n {{{\tan }^{ - 1}}\left( {{1 \over {{r^2} + 3r + 3}}} \right)} } \right\}n→∞lim6tan{r=1∑ntan−1(r2+3r+31)} is equal to :AOption A. 1BOption B. 2COption C. 3DOption D. 6CheckSkip · show solution
