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How can I evaluate this limit? $$ \lim_{x\to 0} \frac{\sin(x) + \sin(x + \pi/N) + \sin(x + 2\pi/N)+...+\sin(x + 2N\pi/N)}{x} $$

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  • $\begingroup$ Use $\sin A=(1/2i)(e^{\pi iA}-e^{-\pi iA})$. $\endgroup$ – Gerry Myerson Nov 16 '17 at 21:47
  • $\begingroup$ I got $1$. Is it really? $\endgroup$ – Michael Rozenberg Nov 16 '17 at 21:51
  • $\begingroup$ Yes. The expression inside the limit is exactly the same as $\frac{\sin x}{x}$, so the limit is $1$. $\endgroup$ – Sangchul Lee Nov 16 '17 at 21:51
  • $\begingroup$ You can check with L'Hopital, but it should be $1$. $\endgroup$ – aleden Nov 16 '17 at 21:52
  • $\begingroup$ @aleden Why does the numerator $\to 0?$ $\endgroup$ – zhw. Nov 17 '17 at 0:02
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For each fixed $x,$ the numerator is the imaginary part of

$$e^{ix} + e^{i(x+ \pi/N)} + e^{i(x+ 2(\pi/N))}\cdots + e^{i(x+ 2N(\pi/N)} = e^{ix}(1+e^{i\pi/N} + (e^{ i\pi/N})^2 + \cdots + (e^{ i\pi/N})^{2N}) = e^{ix}\frac{(e^{ i\pi/N})^{2N+1} -1 }{e^{ i\pi/N}-1}= e^{ix}.$$

The imaginary part of the last term is of course $\sin x.$ Thus the desired limit is $\lim_{x\to 0}(\sin x)/x =1.$

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Multiply both numerator and denominator by $2\sin(\pi/2N)$, and use $2\sin x\sin(\pi/2N)=\cos(x-\pi/2N)-\cos(x+\pi/2N)$ and the numerator can be dealt with telescoping method to get $[\cos(x-\pi/2N)-\cos(x+2\pi/N)]/[2x\sin(\pi/2N)]=\sin x/x\rightarrow 1$ as $x\rightarrow 0$.

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