lim _{n rightarrow infty} frac{pi}{2 n}left[s | vedclass

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(A) The given limit is of the form $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} f\left(\frac{r}{n}\right) = \int_{0}^{1} f(x) dx$. We can

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(A) The given limit is of the form $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} f\left(\frac{r}{n}\right) = \int_{0}^{1} f(x) dx$. We can rewrite the expression as: $\lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{\pi}{2 n} \sin \left( r \cdot \frac{\pi}{2 n} \right)$. Let $x = \frac{r \pi}{2 n}$,then $dx = \frac{\pi}{2 n}$. As $r=1$,$x \rightarrow 0$ and as $r=n$,$x \rightarrow \frac{\pi}{2}$. The integral becomes: $\int_{0}^{\pi/2} \sin(x) dx$. Evaluating the integral: $[-\cos(x)]_{0}^{\pi/2} = -\cos(\frac{\pi}{2}) - (-\cos(0)) = -0 + 1 = 1$.

$\lim _{n \rightarrow \infty} \frac{\pi}{2 n}\left[\sin \frac{\pi}{2 n}+\sin \frac{2 \pi}{2 n}+\sin \frac{3 \pi}{2 n}+\ldots+\sin \frac{\pi}{2}\right]=$

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