1. **State the problem:** Find the limit $$\lim_{x \to 3} \frac{\ln(x-2)}{2x^2 - 6x}$$. 2. **Recall the formula and rules:** We are dealing with a limit of a fraction where the numerator is a natural logarithm function and the denominator is a quadratic expression. If direct substitution leads to an indeterminate form like $$\frac{0}{0}$$, we can apply L'Hôpital's Rule. 3. **Evaluate the limit by direct substitution:** Substitute $$x=3$$: $$\ln(3-2) = \ln(1) = 0$$ $$2(3)^2 - 6(3) = 18 - 18 = 0$$ So the limit is of the form $$\frac{0}{0}$$, which is indeterminate. 4. **Apply L'Hôpital's Rule:** Differentiate numerator and denominator separately: - Numerator derivative: $$\frac{d}{dx} \ln(x-2) = \frac{1}{x-2}$$ - Denominator derivative: $$\frac{d}{dx} (2x^2 - 6x) = 4x - 6$$ 5. **Evaluate the new limit:** $$\lim_{x \to 3} \frac{1/(x-2)}{4x - 6} = \lim_{x \to 3} \frac{1}{(x-2)(4x - 6)}$$ Substitute $$x=3$$: $$\frac{1}{(3-2)(4(3) - 6)} = \frac{1}{1 \times (12 - 6)} = \frac{1}{6}$$ 6. **Final answer:** $$\boxed{\frac{1}{6}}$$