Official solution
\[\beta = \mathop {\lim }\limits_{x \to 0} {{\alpha x - ({e^{3x}} - 1)} \over {\alpha x({e^{3x}} - 1)}},\,\alpha \in R\]
\[ = \mathop {\lim }\limits_{x \to 0} {{{\alpha \over 3} - \left( {{{{e^{3x}} - 1} \over {3x}}} \right)} \over {\alpha x\left( {{{{e^{3x - 1}}} \over {3x}}} \right)}}\]
So, \[\alpha = 3\] (to make independent form)
\[\beta = \mathop {\lim }\limits_{x \to 0} {{1 - \left( {{{{e^{3x}} - 1} \over {3x}}} \right)} \over {3x}} = {{1 - {{\left( {3x + {{9{x^2}} \over 2}\, + \,......} \right)} \over {3x}}} \over {3x}}\]
\[ = {{ - \left( {{9 \over 2}{x^2} + {{{{(3x)}^3}} \over {31}}\, + \,....} \right)} \over {9{x^2}}} = {{ - 1} \over 2}\]
\[\therefore\] \[\alpha + \beta = 3 - {1 \over 2} = {5 \over 2}\]