If $\underset{x\to <!--\; \to \; -->0}{lim}\frac{3+\mathrm{\alpha <!--\; \alpha \; -->}\sin <!--\; \; -->x+\mathrm{\beta <!--\; \beta \; -->}\cos <!--\; \; -->x+{\log }_e<!--\; \; -->(1-<!--\; -\; -->x)}{3{\tan }^2<!--\; \; -->x}=\frac{1}{3}$, then 2α − β is equal to

The correct answer is 5.

We are given the following limit equation:
$\underset{x\to 0}{lim}\frac{3+\alpha \sin x+\beta \cos x+{\log }_e(1-x)}{3{\tan }^{2}x}=\frac{1}{3}$

To evaluate this limit, we can write the limit expression by separating the standard limit component for the denominator. Since $\underset{x\to 0}{lim}\frac{{\tan }^{2}x}{x^2}=1$, we can rewrite the expression as:
$\underset{x\to 0}{lim}\frac{3+\alpha \sin x+\beta \cos x+{\log }_e(1-x)}{3x^2}\times \frac{x^2}{{\tan }^{2}x}=\frac{1}{3}$

Using the fact that $\underset{x\to 0}{lim}\frac{x^2}{{\tan }^{2}x}=1$, the limit simplifies to:
$\underset{x\to 0}{lim}\frac{3+\alpha \sin x+\beta \cos x+{\log }_e(1-x)}{3x^2}=\frac{1}{3}$

Now we substitute the Maclaurin series expansions of $\sin x$, $\cos x$, and ${\log }_e(1-x)$ around $x=0$:
$\sin x=x-\frac{x^3}{3!}+\dots$
$\cos x=1-\frac{x^2}{2!}+\dots$
${\log }_e(1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\dots$

Substituting these series into the numerator:
$\underset{x\to 0}{lim}\frac{3+\alpha [x-\frac{x^3}{3!}+\dots ]+\beta [1-\frac{x^2}{2!}+\dots ]+(-x-\frac{x^2}{2}-\dots )}{3x^2}=\frac{1}{3}$

Grouping the terms of the numerator by powers of $x$:
$\underset{x\to 0}{lim}\frac{(3+\beta )+(\alpha -1)x+(-\frac{\beta }{2}-\frac{1}{2})x^2+O\left(x^3\right)}{3x^2}=\frac{1}{3}$

For the limit to exist and be a finite non-zero value, the coefficients of the terms with powers of $x$ strictly less than the power of $x$ in the denominator (which is $x^2$) must equal zero. Thus, we have:
1) Constant term:
$3+\beta =0\Rightarrow \beta =-3$
2) Coefficient of $x$:
$\alpha -1=0\Rightarrow \alpha =1$

Let us verify this by computing the limit value with these coefficients:
$\underset{x\to 0}{lim}\frac{(-\frac{\beta }{2}-\frac{1}{2})x^2}{3x^2}=\frac{-\frac{\beta }{2}-\; \backslash frac\{1\}\{2\}}{3}$
Substituting $\beta =-3$ into this expression yields:
$\frac{-\frac{-3}{2}-\frac{1}{2}}{3}=\frac{\frac{3}{2}-\frac{1}{2}}{3}=\frac{1}{3}$
This perfectly matches the given limit value of $\frac{1}{3}$.

Now we calculate the value of the required expression $2\alpha -\beta$:
$2\alpha -\beta =2\left(1\right)-(-3)=2+3=5$