Let   $\beta =\underset{x\text{}\to 0}{\lim }\frac{\alpha x-(e^{3x}-1)}{\alpha x(e^{3x}-1)}$  for  some $\alpha \in R$  . Then  the value of $\alpha +\beta$   is 

$$\begin{gathered} \beta =\underset{x\to 0}{\text{Lt}}\frac{\alpha x-(e^{3x}-1)}{\alpha x(e^{3x}-1)} \\ =\underset{x\to 0}{\text{Lt}}\frac{\alpha x-(\frac{3x}{1}+\frac{{\left(3x\right)}^2}{2!}+\frac{{\left(3x\right)}^3}{3!}+\mathrm{.......})}{\alpha x(\frac{3x}{1}+\frac{{\left(3x\right)}^2}{2!}+\mathrm{.....})} \\ \alpha -3=0\Rightarrow \alpha =3 \beta =\underset{x\to 0}{Lt}\frac{-\frac{9x^2}{2}}{3\alpha x^2}=\frac{-1}{2} \end{gathered}$$