lim _{n} {rightarrow infty} frac{1}{n} left[ | vedclass

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(C) The given expression is $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n \frac{r}{n} \sin ^{-1} \left( \frac{r}{n} \right)$.This is a Rieman

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$\frac{\pi}{8}$

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(C) The given expression is $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n \frac{r}{n} \sin ^{-1} \left( \frac{r}{n} \right)$.This is a Riemann sum,which can be written as the definite integral $\int_0^1 x \sin ^{-1} x \, dx$.Using integration by parts,let $u = \sin ^{-1} x$ and $dv = x \, dx$. Then $du = \frac{1}{\sqrt{1-x^2}} \, dx$ and $v = \frac{x^2}{2}$.$\int_0^1 x \sin ^{-1} x \, dx = \left[ \frac{x^2}{2} \sin ^{-1} x \right]_0^1 - \int_0^1 \frac{x^2}{2 \sqrt{1-x^2}} \, dx$.Evaluating the first term: $\left[ \frac{1^2}{2} \sin ^{-1}(1) - 0 \right] = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}$.For the second term: $-\frac{1}{2} \int_0^1 \frac{x^2}{\sqrt{1-x^2}} \, dx = -\frac{1}{2} \int_0^1 \frac{x^2 - 1 + 1}{\sqrt{1-x^2}} \, dx = -\frac{1}{2} \left[ \int_0^1 -\sqrt{1-x^2} \, dx + \int_0^1 \frac{1}{\sqrt{1-x^2}} \, dx \right]$.$= \frac{1}{2} \int_0^1 \sqrt{1-x^2} \, dx - \frac{1}{2} \int_0^1 \frac{1}{\sqrt{1-x^2}} \, dx$.Using standard integrals: $\int \sqrt{1-x^2} \, dx = \frac{x}{2} \sqrt{1-x^2} + \frac{1}{2} \sin ^{-1} x$ and $\int \frac{1}{\sqrt{1-x^2}} \, dx = \sin ^{-1} x$.$= \frac{1}{2} \left[ \frac{x}{2} \sqrt{1-x^2} + \frac{1}{2} \sin ^{-1} x \right]_0^1 - \frac{1}{2} \left[ \sin ^{-1} x \right]_0^1$.$= \frac{1}{2} \left[ 0 + \frac{1}{2} \cdot \frac{\pi}{2} - 0 \right] - \frac{1}{2} \left[ \frac{\pi}{2} - 0 \right] = \frac{\pi}{8} - \frac{\pi}{4} = -\frac{\pi}{8}$.Adding the first term: $\frac{\pi}{4} - \frac{\pi}{8} = \frac{\pi}{8}$.

$\lim _{n}$ ${\rightarrow \infty} \frac{1}{n} \left[ \frac{1}{n} \sin ^{-1} \frac{1}{n} + \frac{2}{n} \sin ^{-1} \frac{2}{n} + \dots + \frac{n}{n} \sin ^{-1} \frac{n}{n} \right] =$

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