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

Question

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 1

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

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

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

Answer 2

Let $y=x-\pi/3$, so we are to compute \begin{align*} \lim_{y\rightarrow 0}\dfrac{\tan^{3}(y+\pi/3)-3\tan(y+\pi/3)}{\cos(y+\pi/2)}. \end{align*} Note that \begin{align*} \tan^{3}(y+\pi/3)-3\tan(y+\pi/3)&=\tan(3(y+\pi/3))(3\tan^{2}(y+\pi/3)-1)\\ &=(\tan(3y))(3\tan^{2}(y+\pi/3)-1), \end{align*} and the term reduces to \begin{align*} \dfrac{(\tan(3y))(3\tan^{2}(y+\pi/3)-1)}{-\sin y}. \end{align*} But \begin{align*} \lim_{y\rightarrow 0}\dfrac{\tan 3y}{\sin y}=3\lim_{y\rightarrow 0}\dfrac{\sin 3y}{3y}\dfrac{1}{\cos 3y}\dfrac{y}{\sin y}=3. \end{align*}

Answer 3

$$\begin{aligned}\frac{\tan^3 x-3\tan x}{\cos\left(x+\frac{\pi}{6}\right)} &=\tan(x)\cdot\frac{\tan^2 x-3}{\cos x\cos\left(\frac{\pi}{6}\right)-\sin x\sin\left(\frac {\pi}{6}\right)},\quad\quad x\neq \frac{\pi}{3} \\ &=\frac{2\sin x}{\cos^{3} x}\cdot\frac{\sin x^{2}-3\cos x^{2}}{\sqrt{3}\cos x-\sin x}\quad\quad\text{(1)} \\ &=\frac{2\sin x}{\cos^{3} x}\cdot\frac{\left(\sin x^{2}-3\cos x^{2}\right)\left(\sqrt{3}\cos x+\sin x\right)}{\left(\sqrt{3}\cos x-\sin x\right)\left(\sqrt{3}\cos x+\sin x\right)} \\ &=-\frac{2\sin x}{\cos^{3} x}\cdot\frac{4\cos^{2} x-1}{4\cos^{2} x-1}\cdot\left(\sqrt{3}\cos x+\sin x\right)\quad\quad\text{(2)} \end{aligned}$$

Equation $(1)$ uses $\tan x=\frac{\sin x}{\cos x}$ and $(2)$ uses $\sin^2 x=1-\cos^2 x$. Hence, by substitution, the limit is $-24$.