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 Option 3: 5.

We are given:

$\underset{x\to 0}{lim}\frac{3+\alpha sinx+\beta cosx+{log}_e(1-x)}{3{tan}^{2}x}=\frac{1}{3}$

and we need to find 2α - β.

Step 1: Apply the condition for the limit to exist.

As $x\to 0$, the denominator $3{tan}^{2}x\to 0$. For the overall limit to be a finite value (1/3), the numerator must also tend to 0. Substituting x = 0 in the numerator:

$3+\alpha ·0+\beta ·1+{log}_e\left(1\right)=3+\beta +0=0$

This gives us: β = -3.

Step 2: Expand all terms using Taylor series around x = 0.

Recall the standard expansions:

sin x = x - x³/6 + ...
cos x = 1 - x²/2 + x⁴/24 - ...
log_e(1 - x) = -x - x²/2 - x³/3 - ...
tan²x = x² + (2/3)x⁴ + ...   ⟹   3 tan²x = 3x² + ...

Step 3: Substitute β = -3 and expand the numerator.

Numerator = 3 + α(x - x³/6 + ...) + (-3)(1 - x²/2 + ...) + (-x - x²/2 - ...)

Collecting terms by power of x:

Constant term (x⁰): 3 + (-3) = 0 ✓
Linear term (x¹): αx - x = (α - 1)x
Quadratic term (x²): $+\frac{3}{2}{x}^2-\frac{1}{2}{x}^2=x^2$

Step 4: For the limit to be finite, the x¹ coefficient must vanish.

Since the denominator behaves as 3x² (order x²), if the numerator has an x¹ term, the limit would blow up to ±∞. So we need:

$\alpha -1=0⟹\alpha =1$

Step 5: Verify using the limit condition.

With α = 1 and β = -3, the numerator ~ x² and the denominator ~ 3x², so:

$\underset{x\to 0}{lim}\frac{x^2}{3x^2}=\frac{1}{3}$ ✓

This confirms α = 1 and β = -3.

Step 6: Compute 2α - β.

$2\alpha -\beta =2\left(1\right)-\left(-3\right)=2+3=5$

Therefore, 2α - β = 5.