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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 .

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  • $\begingroup$ any thoughts by yourself? $\endgroup$ – Arnaldo May 17 '17 at 13:14
  • $\begingroup$ @Arnaldo .$\tan 6x+\tan(\frac{\pi}{3}-2x)-\tan(\frac{\pi}{3}+4x))=\tan 6x-\cot(\frac{\pi}{6}-4x)+\cot(\frac{\pi}{6}+2x))$ now ? $\endgroup$ – Almot1960 May 17 '17 at 13:18
  • $\begingroup$ I believe the answer should be approximately 2.33 $\endgroup$ – John Lou May 17 '17 at 13:20
  • $\begingroup$ If you can solve $\lim_{x\to 0}\frac{1}{x^2}(1-\frac{\sin x}{x})$ (with allowed methods) than I will show you the (elementary) rest. :-) $\endgroup$ – user90369 May 17 '17 at 14:29
  • $\begingroup$ @user90369 . how ? please write $\endgroup$ – Almot1960 May 17 '17 at 14:36
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If you know, that $\enspace\displaystyle \lim\limits_{x\to 0}\frac{1}{x^2}(1-\frac{\sin x}{x})=\frac{1}{3!}\enspace$ then you can answer your question easily:

$\displaystyle \frac{(\sin(2x)-2x\cos x)(\tan(6x)+\tan(\frac{\pi}{3}-2x)-\tan(\frac{\pi}{3}+4x))}{x\sin x\tan x\sin(2x)}=$

$\displaystyle =\frac{(\sin(2x)-2x\cos x)(\frac{\sin(6x)}{\cos(6x)}-\frac{\sin(6x)}{\cos(\frac{\pi}{3}-2x)\cos(\frac{\pi}{3}+4x)})}{x\sin x\tan x\sin(2x)}$

$\displaystyle =\frac{2\sin x\cos x -2x\cos x}{\sin x\tan x\sin(2x)}6\frac{\sin(6x)}{6x}(\frac{1}{\cos(6x)}-\frac{1}{\cos(\frac{\pi}{3}-2x)\cos(\frac{\pi}{3}+4x)})$

$\displaystyle =-\frac{1}{x^2}(1-\frac{\sin x}{x}) (\frac{x}{\sin x}\cos x)^2 \frac{2x}{\sin(2x)} 6\frac{\sin(6x)}{6x}(\frac{1}{\cos(6x)}-\frac{1}{\cos(\frac{\pi}{3}-2x)\cos(\frac{\pi}{3}+4x)})$

$\displaystyle \to -\frac{1}{3!}6(1-4)=3\enspace$ for $\enspace x\to 0$

A note about what I have used:

$\displaystyle \tan x=\frac{\sin x}{\cos x}$

$\sin(2x)=2\sin x\cos x$

$\displaystyle \tan x-\tan y=\frac{\sin(x-y)}{\cos x\cos y}$

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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 $$\lim_{x \to 0} \frac{(\sin 2x-2x\cos x)(\tan 6x+\tan(\frac{\pi}{3}-2x)-\tan(\frac{\pi}{3}+4x))}{x\sin x \tan x\sin 2x}=\lim_{x \to 0}\frac{\frac{18}{3}x^4+o(x^5)}{2x^4+o(x^5)}=3. $$

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