What is $\lim\limits_{i\to 0} \frac{2^n}{\frac{(n+1)\sin((n+1)\theta)}{\sin\theta } - \frac{(n-1)\sin((n-1)\theta)}{\sin\theta }} $? What is $$\lim\limits_{i\to 0} \dfrac{2^n}{\frac{(n+1)\sin((n+1)\theta)}{\sin\theta } - \frac{(n-1)\sin((n-1)\theta)}{\sin\theta }} $$ where $$\theta=\frac{i\pi}{n} $$ The second page of this document says, that it's $$\frac{2^{n-2}}{n}$$ but I'm at a complete loss as to how to come to this conclusion.

It's not too hard to show that
$$ \lim_{x \to 0} \frac{\sin{mx}}{nx} = \frac{m}{n}$$
Thus
$$\lim\limits_{i\to 0} \dfrac{2^n}{\frac{(n+1)\sin((n+1)\theta)}{\sin\theta } - \frac{(n-1)\sin((n-1)\theta)}{\sin\theta }} = \frac{2^n}{(n+1)^2-(n-1)^2} = \frac{2^n}{4n} = \frac{2^{n-2}}{n}$$
