JEE MainMathematicsModerateLimits

$\ If\ \lim_{n \rightarrow \infty}\mspace{2mu}\frac{\sum_{r = 1}^{2n}\mspace{2mu}\left( \sqrt{r}\sum_{r = 1}^{2n}\mspace{2mu}\frac{1}{\sqrt{r}} \right)}{4\sum_{r = 1}^{n}\mspace{2mu} r} = \frac{\lambda}{3},\ then\ \lambda =$

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\ \ \lim_{n \rightarrow \infty}\mspace{2mu}\frac{\sum_{r = 1}^{2n}\mspace{2mu}\left( \sqrt{r}\sum_{r = 1}^{2n}\mspace{2mu}\frac{1}{\sqrt{r}} \right)}{4\sum_{r = 1}^{n}\mspace{2mu} r} = \lim_{n \rightarrow \infty}\mspace{2mu}\frac{\sum_{r = 1}^{2n}\mspace{2mu}\sqrt{r}\sum_{r = 1}^{2n}\mspace{2mu}\frac{1}{\sqrt{r}}}{2n(n + 1)} = \lim_{n \rightarrow \infty}\mspace{2mu}\frac{\frac{1}{n}\sum_{r = 1}^{2n}\mspace{2mu}\sqrt{\frac{r}{n}}\frac{1}{n}\sum_{r = 1}^{2n}\mspace{2mu}\frac{1}{\sqrt{\frac{r}{n}}}}{2\left( 1 + \frac{1}{n} \right)}

= \frac{\int_{0}^{2}\mspace{2mu}\sqrt{x}dx\int_{0}^{2}\mspace{2mu}\frac{1}{\sqrt{x}}dx}{2} = \frac{\frac{2}{3}2^{\frac{3}{2}} \cdot 2 \cdot 2^{\frac{1}{2}}}{2} = \frac{8}{3}

Source: Grass-G / M-11 Mathematics Paper-2 1Nov.docx

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