Let $$\beta = \displaystyle\lim_{x \to 0} \dfrac{\alpha x - (e^{3x} - 1)}{\alpha x(e^{3x} - 1)}$$ for some $$\alpha \in \mathbb{R}$$. Then the value of $$\alpha + \beta$$ is:

We need to find $$\alpha + \beta$$ where $$\beta = \displaystyle\lim_{x \to 0} \dfrac{\alpha x - (e^{3x} - 1)}{\alpha x(e^{3x} - 1)}$$ for some $$\alpha \in \mathbb{R}$$.

As $$x \to 0$$: $$e^{3x} - 1 \to 0$$ and $$\alpha x \to 0$$.

The denominator $$\alpha x(e^{3x}-1) \to 0$$.

For the limit to be finite, the numerator must also approach 0:

$$\lim_{x \to 0} [\alpha x - (e^{3x} - 1)] = 0$$

$$\alpha \cdot 0 - (1 - 1) = 0$$. This is $$0$$ regardless of $$\alpha$$.

So we have a $$\frac{0}{0}$$ form. We use Taylor expansion.

$$e^{3x} = 1 + 3x + \frac{(3x)^2}{2} + \frac{(3x)^3}{6} + \cdots = 1 + 3x + \frac{9x^2}{2} + \frac{27x^3}{6} + \cdots$$

$$e^{3x} - 1 = 3x + \frac{9x^2}{2} + O(x^3)$$

$$\alpha x - (e^{3x} - 1) = \alpha x - 3x - \frac{9x^2}{2} - O(x^3) = (\alpha - 3)x - \frac{9x^2}{2} - O(x^3)$$

If $$\alpha \neq 3$$, the numerator is $$O(x)$$ and the denominator is $$O(x^2)$$, giving an infinite limit.

So we need $$\alpha = 3$$.

Numerator: $$-\frac{9x^2}{2} - O(x^3)$$

Denominator: $$3x(3x + \frac{9x^2}{2} + \cdots) = 9x^2 + \frac{27x^3}{2} + \cdots$$

$$\beta = \lim_{x \to 0} \frac{-\frac{9x^2}{2}}{9x^2} = \frac{-9/2}{9} = -\frac{1}{2}$$

$$\alpha + \beta = 3 + \left(-\frac{1}{2}\right) = \frac{5}{2}$$

Therefore, the correct answer is Option C: $$\dfrac{5}{2}$$.