How to decide limits of integral in Riemann's sum

Question

I need to calculate $$\lim_{n \to \infty}\frac{1}{\sqrt{n^2 -0^2}}+\frac{1}{ \sqrt{n^2 -1^2}} + \frac{1}{ \sqrt{n^2 -+2^2}}+.....+ \frac{1}{ \sqrt{n^2 -(n-1)^2} }$$ It can be written as $$ \frac{1}{n}\sum_{r=0}^{n-1}{ \frac{1}{\sqrt{1-(r^2/n^2)}} }$$ and in integral form as $$ \int_?^? \frac{1}{ \sqrt{1-(x^2) }}dx $$ How to decide limits?

Answer 1

The evaluation of the limit of the sum uses lower Riemann sum where each $x_i = \dfrac{i}{n}, i = \overline{0,n-1}$. Thus: $\displaystyle \lim_{n\to \infty} S_n = \displaystyle \int_{0}^1 \dfrac{1}{\sqrt{1-x^2}}dx = \arcsin x|_{x=0}^1 = \dfrac{\pi}{2}$

Answer 2

Consider a limit($\lim_{n\to\infty}$) for limits: $$\frac{1}{n}\sum_{r=0}^{n-1}{ \frac{1}{\sqrt{1-(r^2/n^2)}} }=\int_{\lim_{n\to\infty}0/n}^{\lim_{n\to\infty}(n-1)/n}\frac1{\sqrt{1-(x^2)}}{\rm d}x=\int_{0}^{1}\frac1{\sqrt{1-x^2}}{\rm d}x=\arcsin x|_0^1=\frac{\pi}2$$