The value of lim _{n rightarrow infty} frac{1 | vedclass

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(B) Given limit is $L = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=0}^{2 n-1} \frac{n^{2}}{n^{2}+4 r^{2}}$.Dividing the numerator and denominato

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$\frac{1}{2} 	an ^{-1}(4)$

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(B) Given limit is $L = \lim _{n ightarrow \infty} \frac{1}{n} \sum_{r=0}^{2 n-1} \frac{n^{2}}{n^{2}+4 r^{2}}$.Dividing the numerator and denominator of the fraction by $n^2$,we get:$L = \lim _{n ightarrow \infty} \frac{1}{n} \sum_{r=0}^{2 n-1} \frac{1}{1+4(\frac{r}{n})^2}$.Using the definition of the definite integral as the limit of a sum,$\lim _{n ightarrow \infty} \frac{1}{n} \sum_{r=0}^{kn-1} f(\frac{r}{n}) = \int_{0}^{k} f(x) dx$.Here,$f(x) = \frac{1}{1+4x^2}$ and the upper limit is $k = \lim_{n ightarrow \infty} \frac{2n-1}{n} = 2$.Thus,$L = \int_{0}^{2} \frac{1}{1+(2x)^2} dx$.Using the formula $\int \frac{1}{1+a^2x^2} dx = \frac{1}{a} 	an^{-1}(ax) + C$,we get:$L = \left[ \frac{1}{2} 	an^{-1}(2x) ight]_{0}^{2}$.$L = \frac{1}{2} 	an^{-1}(2 	imes 2) - \frac{1}{2} 	an^{-1}(2 	imes 0)$.$L = \frac{1}{2} 	an^{-1}(4) - 0 = \frac{1}{2} 	an^{-1}(4)$.

The value of $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=0}^{2 n-1} \frac{n^{2}}{n^{2}+4 r^{2}}$ is:

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