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?
 A: 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*}
A: 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:
$$\frac{\sin x+\sin 3x +\sin 5x}{\tan 2x +\tan 4x +\tan 6x}=\frac{9x+o(x)}{12x+o(x)}=\frac{9+o(1)}{12+o(1)}\to \frac34$$
A: $$\lim_{x \to 0} {\sin(x)+\sin(3x)+\sin(5x) \over \tan(2x)+\tan(4x)+\tan(6x)} = \lim_{x \to 0} {\cos(x)+3\cos(3x)+5\cos(5x) \over 2\sec^2(2x)+4\sec^2(4x)+6\sec^2(6x)} = \frac{9}{12} = \frac{3}{4}$$
I used L'Hopital's rule
A: This is a much simpler take on this question and it uses the following result $$\lim_{x\to 0}\sin x = 0\tag{1}$$ from which we get $$\lim_{x \to 0}\cos x = 1\tag{2}$$ using the relation $\sin^{2}x + \cos^{2}x = 1$. Further dividing $(1)$ by $(2)$ we get $$\lim_{x \to 0}\tan x = 0\tag{3}$$ We can see that the numerator $f(x)$ of the given expression can be simplified as
\begin{align}
f(x) &= \sin x + \sin 3x + \sin 5x\notag\\
&= \sin x + \sin 5x + \sin 3x\notag\\
&= 2\sin 3x\cos 2x + \sin 3x\notag\\
&= \sin 3x(2\cos 2x + 1)\notag
\end{align}
and the denominator $g(x)$ can be simplified as
\begin{align}
g(x) &= \tan 2x + \tan 4x + \tan 6x\notag\\
&= \tan 6x(1 - \tan 2x\tan 4x) + \tan 6x\notag\\
&= \tan 6x(2 - \tan 2x\tan 4x)\notag\\
\end{align} In the above we have used the formula $$\tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a\tan b}$$ which can be rewritten as $$\tan a + \tan b = \tan(a + b)(1 - \tan a\tan b)$$
Hence the desired limit $L$ can be calculated as follows
\begin{align}
L &= \lim_{x \to 0}\frac{f(x)}{g(x)}\notag\\
&= \lim_{x \to 0}\frac{\sin 3x(2\cos 2x + 1)}{\tan 6x(2 - \tan 2x\tan 4x)}\notag\\
&= \lim_{x \to 0}\frac{\sin 3x}{\tan 6x}\cdot\lim_{x \to 0}\frac{2\cos 2x + 1}{2 - \tan 2x\tan 4x}\notag\\
&= \lim_{x \to 0}\frac{\sin 3x (1 - \tan^{2}3x)}{2\tan 3x}\cdot\frac{2\cdot 1 + 1}{2 - 0\cdot 0}\notag\\
&= \frac{3}{4}\lim_{x \to 0}\frac{\sin 3x}{\tan 3x}\cdot\lim_{x \to 0}(1 - \tan^{2}3x)\notag\\
&= \frac{3}{4}\lim_{x \to 0}\cos 3x\cdot (1 - 0\cdot 0)\notag\\
&= \frac{3}{4}\cdot 1 = \frac{3}{4}\notag
\end{align}
The calculation above has been presented in somewhat unnecessarily detailed fashion to highlight the fact that no other limits other than $(1), (2), (3)$ are used in the process and as mentioned before out of these the result $(1)$ is the fundamental one. Thus there is no need to invoke the slightly more powerful result $\lim\limits_{x \to 0}\dfrac{\sin x}{x} = 1$.
A: Note that as $x \to 0$, the numerator goes as $x + 3x + 5x$ at leading order, whereas the denominator behaves like $2x+4x+6x$, for which we have: $L =(1+3+5)/(2+4+6) = 3/4$. 
I have used the well-known property $\sin{x} \sim x$ as $x \to 0$.
Hope you find this useful.
