limits - how to find $lim_{large x to frac{pi}{3}}frac{tan^{3}left(xright)-3tanleft(xright)}{cosleft(x+frac{pi}{6}right)}$

how to find

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

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

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

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