Let $$g(x)$$ be a linear function and $$f(x)=\left\{\begin{array}{cl}g(x) & , x \leq 0 \\ \left(\frac{1+x}{2+x}\right)^{\frac{1}{x}} & , x>0\end{array}\right.$$, is continuous at $$x=0$$. If $$f^{\prime}(1)=f(-1)$$, then the value $$g(3)$$ is

Consider the function $$f:(0,2) \rightarrow \mathbf{R}$$ defined by $$f(x)=\frac{x}{2}+\frac{2}{x}$$ and the function $$g(x)$$ defined by

$$g(x)=\left\{\begin{array}{ll} \min \lfloor f(t)\}, & 0<\mathrm{t} \leq x \text { and } 0 < x \leq 1 \\ \frac{3}{2}+x, & 1 < x < 2 \end{array} .\right. \text { Then, }$$

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