If $f\left(x\right)=\{\begin{array}{l}3{(1+\left|\tan x\right|)}^{\frac{\alpha }{\left|\tan x\right|}},\frac{-1}{2}<x<0\\ \beta ,x=0\\ 3{[1+\left|\frac{\sin x}{3}\right|]}^{\frac{2}{\left|\tan x\right|}},0<x<\frac{2}{3}\end{array}$ is a continuous function at $x=0,$ then which of the following Options is/are correct ?

$LHL=f\left(0^{-1}\right)=\underset{x\to 0^{-}}{\lim }3{(1+\left|\tan x\right|)}^{\frac{\alpha }{\left|\tan x\right|}},\left(1^{\infty }form\right)\Rightarrow 3.e^{{ }^{\underset{x\to 0^{-}}{\lim }\frac{\left|\tan x\right|\infty }{\left|\tan x\right|}}}=3e^{\infty }$

Now, $RHL=f\left(0^{+}\right)=\underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }3{[1+\frac{\left|\sin x\right|}{3}]}^{\frac{2}{\left|\tan x\right|}},\left(1^{\infty }form\right)=3e^{\underset{x\to 0^{+}}{\lim }\left|\frac{\sin x}{2}\right|\frac{2}{\left|\tan x\right|}}=3e^{\frac{2}{3}}$

So, for the function $f\left(x\right)$ to be continuous at $x=0$ we must have 

At $x=0,$we must have $3e^{\alpha }=3e^{\frac{2}{3}}=\beta$

$so.\alpha =\frac{2}{3},\beta =3e^{\frac{2}{3}}$