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
$$1+2+\cdots+k = S = \frac{k(k+1)}{2}.$$