Let β = _ x → 0 - ( e ^ 3x - 1) ( e ^ 3x - 1) for some α ∈ ℝ . Then...

Given, $\lim_{x \to 0}\dfrac{\alphax - \left({e}^{3x} - 1\right)}{\alphax\left({e}^{3x} - 1\right)} = \beta$

$⇒ \lim_{x \to 0}\dfrac{\alphax + 1 - {e}^{3x}}{\dfrac{\alphax\left({e}^{3x} - 1\right)}{3x} \cdot 3x} = \beta$

$⇒ \lim_{x \to 0}\dfrac{\alphax + 1 - {e}^{3x}}{3\alpha{x}^{2}} = \beta$ as \left(\lim_{x \to 0}\dfrac{{e}^{x} - 1}{x} = 1\right)$</p><p>$⇒ \lim_{x \to 0}\dfrac{\alpha - 3{e}^{3x}}{6\alphax} = \beta$[Using L' Hospital rule as \left(\dfrac{0}{0}\text{form}\right)$]

Now for limit to exist, $\alpha = 3$

So, $\lim_{x \to 0}\dfrac{3 - 3{e}^{3x}}{6 \times 3 \times x} = \beta$

So, $\beta = \lim_{x \to 0}\dfrac{- \left({e}^{3x} - 1\right)}{6x} = \lim_{x \to 0}\dfrac{- 3{e}^{3x}}{6} = - \dfrac{1}{2}$

$∴ \alpha + \beta = \dfrac{5}{2}$