Math, asked by Anonymous, 1 day ago


 \huge \bold \red {{ \displaystyle \lim_{n \to  \infty}} \frac{1}{n}  \sum\limits_{r=1}^{2n}  \frac{r}{ \sqrt{ {n}^{2}  +  {r}^{2}}}}


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Answers

Answered by sajan6491
14

 \large \bold{\red {{ \displaystyle \lim_{n \to \infty}} \frac{1}{n} \sum\limits_{r=1}^{2n} \frac{r}{ \sqrt{ {n}^{2} + {r}^{2}}}}}

 \large \bold \red{{\red{{{  \implies\displaystyle \lim_{n \to \infty}}\sum\limits_{r=1}^{2n} \frac{ \frac{r}{n} }{ \sqrt{1 + ( \frac{r}{u} ) {}^{2} } } }}}}

  \displaystyle \bold \red{\frac{1}{2}\int_{0}^{2}  \frac{2x}{ \sqrt{1+  {x}^{2}}} =  \frac{1}{2}\int_{1}^{5}m {}^{ \frac{ - 1}{2} } dm}

 \bold \red{ { \displaystyle{=  \frac{1}{2} \frac{m {}^{ \frac{1}{2} } }{ \frac{1}{2} }  \int_{1}^{5} =  \sqrt{5} } - 1}}

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