The value of $$\mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 0}^{2n - 1} {{{{n^2}} \over {{n^2} + 4{r^2}}}} $$ is :
JEE · Math · previous-year question
- A.$${1 \over 2}{\tan ^{ - 1}}(2)$$
- B.$${1 \over 2}{\tan ^{ - 1}}(4)$$correct
- C.$${\tan ^{ - 1}}(4)$$
- D.$${1 \over 4}{\tan ^{ - 1}}(4)$$
Answer
B. $${1 \over 2}{\tan ^{ - 1}}(4)$$
Explanation
$$L = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 0}^{2n - 1} {{1 \over {1 + 4{{\left( {{r \over n}} \right)}^2}}}} $$ $$ \Rightarrow L = \int\limits_0^2 {{1 \over {1 + 4{x^2}}}dx} $$ $$ \Rightarrow L = \left. {{1 \over 2}{{\tan }^{ - 1}}(2x)} \right|_0^2 \Rightarrow L = {1 \over 2}{\tan ^{ - 1}}4$$
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