1. **State the problem:** Find the limit $$\lim_{x \to 0} (1 + 2^x)^{\frac{3}{x}}.$$\n\n2. **Recall the formula:** Limits of the form $$\lim_{x \to 0} (1 + f(x))^{\frac{1}{x}} = e^{\lim_{x \to 0} \frac{f(x)}{x}}$$ if the limit on the right exists.\n\n3. **Rewrite the expression:** Let $$y = (1 + 2^x)^{\frac{3}{x}} = \left[(1 + 2^x)^{\frac{1}{x}}\right]^3.$$\n\n4. **Focus on the inner limit:** Consider $$L = \lim_{x \to 0} (1 + 2^x)^{\frac{1}{x}}.$$\n\n5. **Use logarithms:** $$\ln L = \lim_{x \to 0} \frac{\ln(1 + 2^x)}{x}.$$\n\n6. **Expand $2^x$ near 0:** Using the exponential approximation, $$2^x = e^{x \ln 2} \approx 1 + x \ln 2$$ for small $x$.\n\n7. **Approximate inside the logarithm:** $$1 + 2^x \approx 1 + 1 + x \ln 2 = 2 + x \ln 2.$$\n\n8. **Calculate the logarithm:** $$\ln(1 + 2^x) \approx \ln(2 + x \ln 2) = \ln 2 + \ln\left(1 + \frac{x \ln 2}{2}\right) \approx \ln 2 + \frac{x \ln 2}{2}$$ using $\ln(1 + u) \approx u$ for small $u$.\n\n9. **Divide by $x$:** $$\frac{\ln(1 + 2^x)}{x} \approx \frac{\ln 2 + \frac{x \ln 2}{2}}{x} = \frac{\ln 2}{x} + \frac{\ln 2}{2}.$$\n\n10. **Evaluate the limit:** The term $\frac{\ln 2}{x}$ diverges as $x \to 0$, so we need a more precise approach.\n\n11. **Use a better expansion:** Instead, write $$\ln(1 + 2^x) = \ln(1 + e^{x \ln 2}) = \ln(1 + 1 + x \ln 2 + \frac{(x \ln 2)^2}{2} + \cdots) = \ln(2 + x \ln 2 + \cdots).$$\n\n12. **Rewrite as:** $$\ln(2 + x \ln 2) = \ln 2 + \ln\left(1 + \frac{x \ln 2}{2}\right) \approx \ln 2 + \frac{x \ln 2}{2}.$$\n\n13. **Divide by $x$ again:** $$\frac{\ln(1 + 2^x)}{x} \approx \frac{\ln 2 + \frac{x \ln 2}{2}}{x} = \frac{\ln 2}{x} + \frac{\ln 2}{2}.$$\n\n14. **This suggests divergence, so use L'Hôpital's Rule:** Define $$g(x) = \ln(1 + 2^x), \quad h(x) = x.$$ Then $$\lim_{x \to 0} \frac{g(x)}{h(x)} = \lim_{x \to 0} \frac{g'(x)}{h'(x)}.$$\n\n15. **Compute derivatives:** $$g'(x) = \frac{2^x \ln 2}{1 + 2^x}, \quad h'(x) = 1.$$\n\n16. **Evaluate at $x=0$:** $$g'(0) = \frac{2^0 \ln 2}{1 + 2^0} = \frac{1 \cdot \ln 2}{1 + 1} = \frac{\ln 2}{2}.$$\n\n17. **Therefore:** $$\ln L = \frac{\ln 2}{2} \implies L = e^{\frac{\ln 2}{2}} = 2^{\frac{1}{2}} = \sqrt{2}.$$\n\n18. **Recall the original limit:** $$\lim_{x \to 0} (1 + 2^x)^{\frac{3}{x}} = L^3 = (\sqrt{2})^3 = 2^{\frac{3}{2}} = 2 \sqrt{2}.$$\n\n**Final answer:** $$\boxed{2 \sqrt{2}}.$$