TITLE: Trigonometry limit's proof: $\lim_{x\to0}\frac{\sin(x)+\sin(2x)+\cdots+\sin(kx)}{x}=\frac{k(k+1)}{2}$
QUESTION [10 upvotes]: How to prove that $$\lim_{x\to0}\frac{\sin(x)+\sin(2x)+\cdots+\sin(kx)}{x}=\frac{k(k+1)}{2}$$
I tried to split up the fraction and multiple-divide every new fraction with its $x$ factor but didn't work out.
ex: $$\lim_{x\to 0}\frac{\sin(2x)}{x} = \lim_{x\to 0}\frac{\sin(2x)\cdot 2}{2\cdot x}=2$$

REPLY [2 votes]: Method$\#1:$
If $S=\sum_{r=1}^n\sin(rx),$
Using Prosthaphaeresis Formula,
$$S=\dfrac12\sum_{r=1}^n[\sin(rx)+\sin(n+1-r)x]=\sin\dfrac{(n+1)x}2\sum_{r=1}^n\cos\dfrac{(n+1-2r)x}2$$
Now use $\lim_{h\to0}\dfrac{\sin h}h=\lim_{h\to0}\cos h=1$
Method$\#2:$
Use How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?
and off course  $\lim_{h\to0}\dfrac{\sin h}h=1$