Evaluate the limit of ratio of sums of sines (without L'Hopital): $\lim_{x\to0} \frac{\sin x+\sin3x+\sin5x}{\sin2x+\sin4x+\sin6x}$ Limit to evaluate:
$$\lim_{x \rightarrow 0} \cfrac{\sin{(x)}+\sin{(3x)}+\sin{(5x)}}{\sin{(2x)}+\sin{(4x)}+\sin{(6x)}}$$
Proposed solution:
$$
\cfrac{\sin(x)+\sin(3x)+\sin(5x)}{\sin(2x)+\sin(4x)+\sin(6x)}
\Bigg/ \cdot\ \cfrac{1/x}{1/x}\Bigg/=
\frac{\cfrac{\sin(x)}x + \cfrac{\sin(3x)}{3x} \cdot 3 + \cfrac{\sin(5x)}{5x} \cdot 5}
{\cfrac{\sin(2x)}{2x} \cdot 2 + \cfrac{\sin(4x)}{4x} \cdot 4 + \cfrac{\sin(6x)}{6x} \cdot 6}
$$
Using $\lim_{x \rightarrow 0} \frac{\sin x}x=1$, we get
$$\frac{1+1\cdot 3+1\cdot 5}{1\cdot 2+1\cdot 4+1\cdot 6} = \frac 9{12} = \frac 3 4$$
Please tell me if I am correct. 

Hint:
$${f(x)\over g(x)}={{f(x)\over x}\over{g(x)\over x}}$$
