Tuesday, 18 February 2014

limits - how to find limlargextofracpi3fractan3left(xright)3tanleft(xright)cosleft(x+fracpi6right)



how to find





lim without L'hopital or taylor/Laurent series




I tried but did not get any answer:



\frac{\tan^{2}\left(x\right)\tan\left(x\right)-3\tan\left(x\right)}{\cos\left(x+\frac{\pi}{6}\right)}=\frac{2\left(\sin^{3}\left(x\right)-3\sin\left(x\right)\cos^{2}\left(x\right)\right)}{\cos^{3}\left(x\right)\left(\sqrt{3}\cos\left(x\right)-\sin\left(x\right)\right)}=\frac{2\left(\sin^{3}\left(x\right)-3\left(\sin\left(x\right)\left(1-\sin^{2}\left(x\right)\right)\right)\right)}{\cos^{3}\left(x\right)\left(\sqrt{3}\cos\left(x\right)-\sin\left(x\right)\right)}=\frac{2}{1}\frac{-3\left(\sin(x)-\sin^{2}\left(x\right)\right)+\sin^{3}\left(x\right)}{\left(1-\sin^{2}\left(x\right)\right)(\cos(x))\left(\sqrt{3}\cos\left(x\right)-\sin\left(x\right)\right)}


Answer



\lim_{\large x \to \frac{\pi}{3}}\frac{\tan^{3}\left(x\right)-3\tan\left(x\right)}{\cos\left(x+\frac{\pi}{6}\right)}=




\lim_{\large x \to \frac{\pi}{3}}\frac{\tan x (\tan x - \sqrt 3)(\tan x +\sqrt 3)}{\cos x \cos (\pi /6)-\sin x \sin (\pi /6)}=



\lim_{\large x \to \frac{\pi}{3}}\frac{(2\sec x)\tan x (\tan x - \sqrt 3)(\tan x +\sqrt 3)}{\sqrt 3 - \tan x}=-24


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