fine the limits :$\lim_{x \to 0} \frac{(\sin 2x-2x\cos x)(\tan 6x+\tan(\frac{\pi}{3}-2x)-\tan(\frac{\pi}{3}+4x))}{x\sin x \tan x\sin 2x}=?$ fine the limits-without-lhopital rule and  Taylor series :
$$\lim_{x \to 0} \frac{(\sin 2x-2x\cos x)(\tan 6x+\tan(\frac{\pi}{3}-2x)-\tan(\frac{\pi}{3}+4x))}{x\sin x \tan x\sin 2x}=?$$
i know that :
$$\lim_{x \to 0} \frac{\sin x}{x}=1=\lim_{x \to 0}\frac{\tan x}{x}$$
But I can not answer please help .

Here is a different way. 
You can use standard Taylor series expansions, as $x \to 0$,  to get $$\sin 2x-2x\cos x=-\frac{x^3}3+o(x^4) $$ $$\tan 6x+\tan(\frac{\pi}{3}-2x)-\tan(\frac{\pi}{3}+4x))=-18x+o(x^2) $$ $$x\sin x \tan x\sin 2x=2x^4+o(x^5)$$ then $$\lim_{x \to 0} \frac{(\sin 2x-2x\cos x)(\tan 6x+\tan(\frac{\pi}{3}-2x)-\tan(\frac{\pi}{3}+4x))}{x\sin x \tan x\sin 2x}=\lim_{x \to 0}\frac{\frac{18}{3}x^4+o(x^5)}{2x^4+o(x^5)}=3. $$
