# calculating a limit $\lim_{x\to 0} \frac{sinx + sin(x + \pi/N) + sin(x + 2\pi/N)+…+sin(x + 2N\pi/N)}{x}$

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

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

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 $\lim_{x\to 0}(\sin x)/x =1.$
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 $x\rightarrow 0$.