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?

In what follows I will provide full details of the idea of Abhishek Parab.
First divide top and bottom by $x,$ then
\begin{eqnarray*}
\frac{\sin x+\sin 3x+\sin 5x}{\tan 2x+\tan 4x+\tan 6x} &=&\frac{\left( 
\dfrac{\sin x+\sin 3x+\sin 5x}{x}\right) }{\left( \dfrac{\tan 2x+\tan
4x+\tan 6x}{x}\right) } \\
&& \\
&=&\frac{\left( \dfrac{\sin x}{x}+3\left( \dfrac{\sin 3x}{3x}\right)
+5\left( \dfrac{\sin 5x}{5x}\right) \right) }{\left( 2\left( \dfrac{\tan 2x}{%
2x}\right) +4\left( \dfrac{\tan 4x}{4x}\right) +6\left( \dfrac{\tan 6x}{6x}%
\right) \right) }
\end{eqnarray*}
Now using the standard limits
\begin{equation*}
\lim_{u\rightarrow 0}\frac{\sin u}{u}=\lim_{u\rightarrow 0}\frac{\tan u}{u}=1
\end{equation*}
the limit follows
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{\sin x+\sin 3x+\sin 5x}{\tan 2x+\tan 4x+\tan 6x}
&=&\frac{\left( \lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x}+3\left(
\lim\limits_{3x\rightarrow 0}\dfrac{\sin 3x}{3x}\right) +5\left(
\lim\limits_{5x\rightarrow 0}\dfrac{\sin 5x}{5x}\right) \right) }{\left(
2\left( \lim\limits_{2x\rightarrow 0}\dfrac{\tan 2x}{2x}\right) +4\left(
\lim\limits_{4x\rightarrow 0}\dfrac{\tan 4x}{4x}\right) +6\left(
\lim\limits_{6x\rightarrow 0}\dfrac{\tan 6x}{6x}\right) \right) } \\
&=& \\
&=&\frac{\left( 1+3\left( 1\right) +5\left( 1\right) \right) }{\left(
2\left( 1\right) +4\left( 1\right) +6\left( 1\right) \right) }=\frac{3}{4}.
\end{eqnarray*}
