Need a hint to evaluate $\lim_{x \to 0} {\sin(x)+\sin(3x)+\sin(5x) \over \tan(2x)+\tan(4x)+\tan(6x)}$ I know that $\sin A + \sin B + \sin C = 4\cos({A \over 2})\cos({B \over 2})\cos({C \over 2})$ when $A+B+C=\pi$. If ${x \to 0}$ then I have a half circle, right?
If it is right then I have
$\tan(2x) + \tan(4x) + \tan(6x)=\tan(2x)\tan(4x)\tan(6x)$.
I got stuck at
$${4\cos({x \over 2})\cos({3x \over 2})\cos({5x \over 2}) \over \tan(2x)\tan(4x)\tan(6x)}$$
Is this correct? Do you have a hint to help me?

With Taylor's formula at order $1$:
$\sin x+\sin 3x +\sin 5x= x+3x+5x+o(x)$, $\,\tan 2x +\tan 4x +\tan 6x=2x+4x+6x+o(x)$, whence:
$$\frac{\sin x+\sin 3x +\sin 5x}{\tan 2x +\tan 4x +\tan 6x}=\frac{9x+o(x)}{12x+o(x)}=\frac{9+o(1)}{12+o(1)}\to \frac34$$
