Question

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


❖ᴏɴʟʏ ᴘʀᴏᴘᴇʀ ꜱᴏʟᴠᴇᴅ ᴀɴꜱᴡᴇʀ ᴡɪᴛʜ ɢᴏᴏᴅ ᴇxᴘʟᴀɴᴀɪᴏɴ ɴᴇᴇᴅᴇᴅ
❖ ɴᴏ ꜱᴘᴀᴍᴍɪɴɢ
❖ᴏɴʟʏ ꜰᴏʀ ᴍᴏᴅᴇʀᴀᴛᴏʀꜱ, ʙʀᴀɪɴʟʏ ꜱᴛᴀʀꜱ ᴀɴᴅ ᴏᴛʜᴇʀ ʙᴇꜱᴛ ᴜꜱᴇʀꜱ​​​​​​​​​​​​​

Answer 1

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

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

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

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