Trigonometry limit's proof: $lim_{xto0}frac{sin(x)+sin(2x)+cdots+sin(kx)}{x}=frac{k(k+1)}{2}$

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: $$\lim_{x\to 0}\frac{\sin(2x)}{x} = \lim_{x\to 0}\frac{\sin(2x)\cdot 2}{2\cdot x}=2$$

Answer

You are so close. Note that

$$\begin{align*} \frac{\sin(x)+\sin(2x)+\cdots+\sin(kx)}{x} &= \frac{\sin(x)}{x}+\frac{\sin(2x)}{x}+\cdots+\frac{\sin(kx)}{x} \\ &= \frac{\sin(x)}{x}+2\frac{\sin(2x)}{2x}+\cdots+k\frac{\sin(kx)}{kx} \\ &\to 1 + 2 + \cdots + k \\ &= \frac{k(k+1)}{2} \end{align*}$$
as $x\to0$.

I suspect you may not have been able to finish because you didn't recognize the identity

$$1 + 2 + \cdots + k = \frac{k(k+1)}{2}.$$
This identity has a very cute proof.
Set $S:=1+2+\cdots+k$. Adding
$$\begin{align*} 1 + 2 + \cdots + k &= S \\ k + (k-1) + \cdots + 1 &= S \\ \end{align*}$$
gives
$$\begin{align*} \underbrace{(k+1)+(k+1)+\cdots+(k+1)}_{k\ \text{times}} = 2S. \\ \end{align*}$$
Therefore $k(k+1)=2S$ and consequently
$$1+2+\cdots+k = S = \frac{k(k+1)}{2}.$$