1. **State the problem:** Find the limit $$\lim_{x \to 0} \frac{e^{3x} - 1}{1 - 8^x}$$. 2. **Recall the formula and rules:** For limits of the form $$\frac{f(x)-f(a)}{g(x)-g(a)}$$ as $$x \to a$$, if direct substitution gives $$\frac{0}{0}$$, use expansions or L'Hôpital's Rule. 3. **Check direct substitution:** $$e^{3\cdot0} - 1 = 1 - 1 = 0$$ $$1 - 8^0 = 1 - 1 = 0$$ So the limit is of indeterminate form $$\frac{0}{0}$$. 4. **Apply L'Hôpital's Rule:** Differentiate numerator and denominator with respect to $$x$$: - Numerator derivative: $$\frac{d}{dx}(e^{3x} - 1) = 3e^{3x}$$ - Denominator derivative: $$\frac{d}{dx}(1 - 8^x) = -8^x \ln(8)$$ 5. **Evaluate the new limit:** $$\lim_{x \to 0} \frac{3e^{3x}}{-8^x \ln(8)} = \frac{3e^0}{-8^0 \ln(8)} = \frac{3}{-1 \cdot \ln(8)} = -\frac{3}{\ln(8)}$$ 6. **Final answer:** $$\boxed{-\frac{3}{\ln(8)}}$$ This means as $$x$$ approaches zero, the expression approaches $$-\frac{3}{\ln(8)}$$.