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

Given, $\underset{x\to 0}{\lim }\frac{\alpha x-\left(e^{3x}-1\right)}{\alpha x\left(e^{3x}-1\right)}=\beta$

$\Rightarrow \underset{x\to 0}{\lim }\frac{\alpha x+1-e^{3x}}{\frac{\alpha x\left(e^{3x}-1\right)}{3x}·3x}=\beta$

$\Rightarrow \underset{x\to 0}{\lim }\frac{\alpha x+1-e^{3x}}{3\alpha x^2}=\beta$ as $\left(\underset{x\to 0}{\lim }\frac{e^x-1}{x}=1\right)$

$\Rightarrow \underset{x\to 0}{\lim }\frac{\alpha -3e^{3x}}{6\alpha x}=\beta$   [Using L' Hospital rule as $\left(\frac{0}{0} \text{form}\right)$]

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

So, $\underset{x\to 0}{\lim }\frac{3-3e^{3x}}{6\times 3\times x}=\beta$

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

$\therefore \alpha +\beta =\frac{5}{2}$