<s>[INST] 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?
 [/INST] 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 $\lim\limits_{x \to 0}\dfrac{\sin x}{x} = 1$.
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