Animated Solution for Mathematics - Limits, Continuity and Differentiability: If ${\lim }_{x\to 0}\frac{3+\alpha \sin x+\beta \cos x+{\log }_e(1-x)}{3{\tan }^{2}x}=\frac{1}{3}$, then $2\alpha -\beta$ is equal to :

The problem presents the following limit:

This expression involves a mix of trigonometric functions, a logarithmic term, and a squared tangent in the denominator. To solve this, we must systematically simplify the components using series expansions.

The first rule of limit evaluation is to simplify the denominator. As $x\to 0$, the function $\tan x$ behaves asymptotically like $x$.

Mathematically, we use the approximation $\tan x\approx x$. Therefore, the denominator $3{\tan }^{2}x$ simplifies to:

This serves as our anchor. It dictates that the numerator must also be of order $x^2$ for the limit to converge to a non-zero finite value.

To analyze the numerator, we employ Maclaurin series expansions. We expand each term up to the $x^2$ power, as higher-order terms will vanish when divided by $x^2$:

Substituting these expansions into the numerator, we obtain:

Grouping the terms by powers of $x$, we get:

For the limit to be finite, the coefficients of the constant term and the $x$ term must be zero. This yields the system:

With $\alpha =1$ and $\beta =-3$, we verify the coefficient of the $x^2$ term:

The limit becomes:

This confirms our values are correct. Finally, we calculate the requested value: