We start by rewriting the given limit as follows:
lim_(n rarr oo)(1)/(n^(2))sum_(a=0)^(n-1)sqrt(n^(2)-r^(2))\lim _{n \rightarrow \infty} \frac{1}{n^2} \sum_{a=0}^{n-1} \sqrt{n^2-r^2}
Next, we’ll express
(sqrt(n^(2)-r^(2)))/(n^(2))\frac{\sqrt{n^2-r^2}}{n^2} as:
(1)/(n)[sqrt(1-((r)/(n))^(2))]\frac{1}{n}\left[\sqrt{1-\left(\frac{r}{n}\right)^2}\right]
Now, we rewrite the limit as a sum:
{:[lim_(n rarr oo)sum_(r=0)^(n-1)(1)/(n)[sqrt(1-((r)/(n))^(2))]=lim_(n rarr oo)sum_(r=1)^(n)(1)/(n)sqrt(1-((r)/(n))^(2))],[=int_(0)^(1)sqrt(1-x^(2))dx],[=[(xsqrt(1-x^(2)))/(2)+(1)/(2)sin^(-1)x]_(0)^(1)],[=(1)/(2)sin^(-1)(1)]:}\begin{aligned}
\lim _{n \rightarrow \infty} \sum_{r=0}^{n-1}\frac{1}{n}\left[\sqrt{1-\left(\frac{r}{n}\right)^2}\right] &= \lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{n} \sqrt{1-\left(\frac{r}{n}\right)^2} \\
& = \int_0^1 \sqrt{1-x^2} dx \\
& = \left[\frac{x \sqrt{1-x^2}}{2}+\frac{1}{2} \sin ^{-1} x\right]_0^1 \\
& = \frac{1}{2} \sin ^{-1}(1)
\end{aligned}
Now, we calculate
sin^(-1)(1)\sin^{-1}(1):
(1)/(2)sin^(-1)(1)=(1)/(2)xx(pi)/(2)=(pi)/(4)\frac{1}{2} \sin ^{-1}(1) = \frac{1}{2} \times \frac{\pi}{2} = \frac{\pi}{4}
Hence, the limit
lim_(n rarr-oo)(1)/(n^(2))sum_(r=0)^(n-1)sqrt(n^(2)-r^(2))\lim _{n \rightarrow-\infty} \frac{1}{n^2} \sum_{r=0}^{n-1} \sqrt{n^2-r^2} is
(pi)/(4)\frac{\pi}{4}.