Evaluate limits of trigonometric equation Evaluate the limit
$$\lim_{x\to0} \frac{\sin x+\sin3x+\sin5x}{\sin2x+\sin4x+\sin6x}$$
I have tried expanding out the equation but it gets really long and tedious.
Is there a better way to solve this question?
 A: Notice, $$\lim_{x\to 0}\frac{\sin x+\sin 3x+\sin 5x}{\sin 2x+\sin 4x+\sin 6x}$$
$$\lim_{x\to 0}\frac{\frac{1}{x}(\sin x+\sin 3x+\sin 5x)}{\frac{1}{x}(\sin 2x+\sin 4x+\sin 6x)}$$
$$\lim_{x\to 0}\frac{\frac{\sin x}{x}+3\frac{\sin 3x}{3x}+5\frac{\sin 5x}{5x}}{2\frac{\sin 2x}{2x}+4\frac{\sin 4x}{4x}+6\frac{\sin 6x}{6x}}$$
Now, using $\lim_{u\to 0}\frac{\sin u}{u}=1$
$$=\frac{1+3+5}{2+4+6}=\frac{9}{12}=\color{red}{\frac{3}{4}}$$
A: $$\begin{align}\lim_{x\to 0}\frac{\sin x+\sin 3x+\sin 5x}{\sin 2x+\sin 4x+\sin 6x}&=\lim_{x\to 0}\frac{\dfrac{\sin x+\sin 3x+\sin 5x}{x}}{\dfrac{\sin 2x+\sin 4x+\sin 6x}{x}}\\\\&=\lim_{x\to 0}\frac{\dfrac{\sin x}{x}+3\cdot \dfrac{\sin 3x}{3x}+5\cdot \dfrac{\sin 5x}{5x}}{2\cdot\dfrac{\sin 2x}{2x}+4\cdot\dfrac{\sin 4x}{4x}+6\cdot\dfrac{\sin 6x}{6x}}\\\\&=\frac{1+3+5}{2+4+6}=\frac 34\end{align}$$
where we use
$\lim_{\circ\to 0}\frac{\sin\circ}{\circ}=1.$
A: Use Taylor at order $1$:
\begin{align*}
\frac{\sin x + \sin 3x + \sin 5x}{\sin 2x + \sin 4x + \sin 6x}&=\frac{x+o(x)+3x+o(x)+5x+o(x)}{2x+o(x)+4x+o(x)+6x+o(x)}=\frac{9x+o(x)}{12x+o(x)}\\&=\frac{3+o(1)}{4+o(1)}\to \frac34.
\end{align*}
A: We know by L'Hospital's Rule
$$ \lim_{x \to 0} \frac{\sin p x }{\sin q x} = \frac{p}{q} $$
So the limit is simply
$$ \dfrac{1+3+5}{2+4+6}=\dfrac34. $$
