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Use Riemann Sums to determine the following limit:

$$\lim_{n \to \infty} \frac 1n \left(\sin \frac \pi n + \sin \frac{2\pi}{n} + \cdots + \sin\frac{(n-1)\pi}{n}\right)$$

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$$\lim_{n \to \infty} \frac 1n \left(\sin \frac \pi n + \sin \frac{2\pi}{n} + \cdots + \sin\frac{(n-1)\pi}{n}\right)=\int_0^1\sin\pi x\,dx=\frac2\pi$$

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HINT: Multiply and divide by $\pi$. Then the term in front of the parenthesis will be $\frac{\pi}{n}$ and that's your $\Delta x$. Now it's easy to spot that it's the Riemann sum for $f(x) = \sin x$ from $0$ to $\pi$.

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Let

$$S_n=\frac{\pi-0}{n}\sum_{k=0}^{n-1}\sin(0+k\frac{\pi-0}{n})$$

the sinus function is continuous at $[0,\pi]$ thus

$lim_{n\to +\infty}S_n=\int_0^\pi sin(t)dt=2$.

and your limit is

$$\color{green}{\frac{2}{\pi}}$$

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