Trigonometry limit's proof: $\lim_{x\to0}\frac{\sin(x)+\sin(2x)+\cdots+\sin(kx)}{x}=\frac{k(k+1)}{2}$ 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: $$\lim_{x\to 0}\frac{\sin(2x)}{x} = \lim_{x\to 0}\frac{\sin(2x)\cdot 2}{2\cdot x}=2$$
 A: This is the derivative at $0$ of the function
$$
f(x)=\sin(x)+\sin(2x)+\dots+\sin(kx)
$$
and
$$
f'(x)=\cos(x)+2\cos(2x)+\dots+k\cos(kx)
$$
Therefore
$$
f'(0)=1+2+\dots+k
$$
and you're done with child Gauss’ trick.
Note: This isn't using l’Hôpital, but the definition of derivative and the chain rule.
A: 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
$$1+2+\cdots+k = S = \frac{k(k+1)}{2}.$$
A: 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  $\lim_{h\to0}\dfrac{\sin h}h=1$
