Limit of limits 2

Calculus Level 3

$\Large \lim _{ n\rightarrow \infty } \frac { 1 }{ n } \sum _{ r=1 }^{ 2n }{ \frac { r }{ \sqrt { { n }^{ 2 }+{ r }^{ 2 } } } } = \ ?$

\sqrt { 2 } -1

\sqrt { 2 } +1

\sqrt { 5 }+1

\sqrt { 5 } -1

1 solution

Rishabh Jain
Apr 8, 2016

The limit is $\LARGE\displaystyle\lim_{n\to\infty}\dfrac{1}{n}\sum_{r=1}^{2n}\dfrac{\frac rn}{\sqrt{1+\frac{r^2}{n^2}}}$

By Reimann Sums , this is:-

$\LARGE \displaystyle\int_0^2\dfrac{x\mathrm{d}x}{\sqrt{1+x^2}}$

Observing $\mathrm{d}(1+x^2)=2x\mathrm{d}x$ , integral simplifies to $\LARGE [\sqrt{1+x^2}]_{0}^2$

$=\huge\boxed{\sqrt 5-1}$