operatorname{Lim}_{n rightarrow infty} frac{p | vedclass

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Question

(A) The given expression is $\lim _{n \rightarrow \infty} \frac{\pi}{2 n}\left[\sin \frac{\pi}{2 n}+\sin \frac{2 \pi}{2 n}+\ldots+\sin \frac{n \pi}{2

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) The given expression is $\lim _{n \rightarrow \infty} \frac{\pi}{2 n}\left[\sin \frac{\pi}{2 n}+\sin \frac{2 \pi}{2 n}+\ldots+\sin \frac{n \pi}{2 n}\right]$.This can be written as $\lim _{n \rightarrow \infty} \frac{\pi}{2 n} \sum_{r=1}^n \sin \left(\frac{r \pi}{2 n}\right)$.Using the definition of definite integral as the limit of a sum, $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n f\left(\frac{r}{n}\right) = \int_0^1 f(x) dx$.Here, we can rewrite the expression as $\int_0^{\pi/2} \sin(x) dx$ by setting $x = \frac{r\pi}{2n}$, where $dx = \frac{\pi}{2n}$.The limits of integration are $x = 0$ when $r=0$ and $x = \frac{\pi}{2}$ when $r=n$.Thus, the integral becomes $\int_0^{\pi/2} \sin(x) dx$.Evaluating the integral: $[-\cos(x)]_0^{\pi/2} = -(\cos(\frac{\pi}{2}) - \cos(0)) = -(0 - 1) = 1$.

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

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