$$\text { If } \lim _\limits{x \rightarrow 0} \frac{3+\alpha \sin x+\beta \cos x+\log _e(1-x)}{3 \tan ^2 x}=\frac{1}{3} \text {, then } 2 \alpha-\beta \text { is equal to : }$$

<p>$$\lim _\limits{x \rightarrow 0} \frac{3+\alpha \sin x+\beta \cos x+\log _e(1-x)}{3 \tan ^2 x}=\frac{1}{3}$$</p> <p>$$\Rightarrow \lim _\limits{x \rightarrow 0} \frac{3+\alpha\left[x-\frac{x^3}{3 !}+\ldots .\right]+\beta\left[1-\frac{x^2}{2 !}+\frac{x^4}{4 !} \ldots .\right]+\left(-x-\frac{x^2}{2}-\frac{x^3}{3} \ldots\right)}{3 \tan ^2 x}=\frac{1}{3}$$</p> <p>$$\Rightarrow \lim _\limits{x \rightarrow 0} \frac{(3+\beta)+(\alpha-1) x+\left(-\frac{1}{2}-\frac{\beta}{2}\right) x^2+\ldots .}{3 x^2} \times \frac{x^2}{\tan ^2 x}=\frac{1}{3}$$</p> <p>$$\begin{aligned} & \Rightarrow \beta+3=0, \alpha-1=0 \text { and } \frac{-\frac{1}{2}-\frac{\beta}{2}}{3}=\frac{1}{3} \\ & \Rightarrow \beta=-3, \alpha=1 \\ & \Rightarrow 2 \alpha-\beta=2+3=5 \end{aligned}$$</p>