1. We need

$$\lim_{x\to 0}\frac{3+\alpha\sin x+\beta\cos x+\ln(1-x)}{3\tan^2 x}=\frac13.$$

Since $\tan^2 x\to 0$ as $x\to 0$, for the limit to be finite, the numerator must vanish up to order $x^2$.

2. Use standard expansions near $x=0$:

$$\sin x=x+O(x^3),\qquad \cos x=1-\frac{x^2}{2}+O(x^4),$$ $$\ln(1-x)=-x-\frac{x^2}{2}+O(x^3),\qquad \tan x=x+O(x^3).$$

Hence

$$\tan^2 x=x^2+O(x^4).$$

So

$$3\tan^2 x=3x^2+O(x^4).$$

3. Expand the numerator:

$$3+\alpha\sin x+\beta\cos x+\ln(1-x)$$ $$=3+\alpha x+\beta\left(1-\frac{x^2}{2}\right)+\left(-x-\frac{x^2}{2}\right)+O(x^3).$$

Thus

$$= (3+\beta) + (\alpha-1)x - \frac{\beta+1}{2}x^2 + O(x^3).$$

4. For the limit to exist and be finite, constant and linear terms must vanish:

• Constant term:

$$3+\beta=0 \implies \beta=-3.$$

• Linear term:

$$\alpha-1=0 \implies \alpha=1.$$

5. Then the quadratic term in the numerator becomes

$$-\frac{\beta+1}{2}x^2=-\frac{-3+1}{2}x^2=x^2.$$

So numerator $=x^2+O(x^3)$.

Therefore,

$$\lim_{x\to 0}\frac{3+\alpha\sin x+\beta\cos x+\ln(1-x)}{3\tan^2 x} =\lim_{x\to 0}\frac{x^2+O(x^3)}{3x^2+O(x^4)}=\frac13,$$

which matches the given condition.

6. Now compute:

$$2\alpha-\beta=2(1)-(-3)=5.$$

Hence the correct option is C.