1. **State the problem:** We want to find the limit as $x$ approaches 0 of the expression $$\frac{e^{3x} - 1 - 3x}{x^2}.$$\n\n2. **Recall the formula and rules:** The exponential function $e^{u}$ can be expanded using its Taylor series around 0 as $$e^{u} = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \cdots.$$\nHere, $u = 3x$.\n\n3. **Apply the Taylor expansion:** Substitute $u=3x$ into the series: $$e^{3x} = 1 + 3x + \frac{(3x)^2}{2} + \frac{(3x)^3}{6} + \cdots = 1 + 3x + \frac{9x^2}{2} + \frac{27x^3}{6} + \cdots.$$\n\n4. **Rewrite the numerator:** $$e^{3x} - 1 - 3x = \left(1 + 3x + \frac{9x^2}{2} + \cdots\right) - 1 - 3x = \frac{9x^2}{2} + \frac{27x^3}{6} + \cdots.$$\n\n5. **Form the fraction:** $$\frac{e^{3x} - 1 - 3x}{x^2} = \frac{\frac{9x^2}{2} + \frac{27x^3}{6} + \cdots}{x^2} = \frac{9x^2}{2x^2} + \frac{27x^3}{6x^2} + \cdots = \frac{9}{2} + \frac{27x}{6} + \cdots.$$\n\n6. **Take the limit as $x \to 0$:** The terms with $x$ vanish, so $$\lim_{x \to 0} \frac{e^{3x} - 1 - 3x}{x^2} = \frac{9}{2}.$$\n\n**Final answer:** $$\boxed{\frac{9}{2}}.$$