1. **State the problem:** Find the limit as $x \to \infty$ of $$\frac{(3^x - 2)^2}{2 \cdot 9^x}$$. 2. **Rewrite the expression:** Note that $9^x = (3^2)^x = 3^{2x}$. So the expression becomes: $$\frac{(3^x - 2)^2}{2 \cdot 3^{2x}}$$ 3. **Expand the numerator:** $$(3^x - 2)^2 = (3^x)^2 - 2 \cdot 3^x \cdot 2 + 2^2 = 3^{2x} - 4 \cdot 3^x + 4$$ 4. **Substitute back:** $$\frac{3^{2x} - 4 \cdot 3^x + 4}{2 \cdot 3^{2x}}$$ 5. **Split the fraction:** $$\frac{3^{2x}}{2 \cdot 3^{2x}} - \frac{4 \cdot 3^x}{2 \cdot 3^{2x}} + \frac{4}{2 \cdot 3^{2x}}$$ 6. **Simplify each term:** $$\frac{3^{2x}}{2 \cdot 3^{2x}} = \frac{\cancel{3^{2x}}}{2 \cdot \cancel{3^{2x}}} = \frac{1}{2}$$ $$\frac{4 \cdot 3^x}{2 \cdot 3^{2x}} = \frac{4}{2} \cdot \frac{3^x}{3^{2x}} = 2 \cdot 3^{-x} = \frac{2}{3^x}$$ $$\frac{4}{2 \cdot 3^{2x}} = 2 \cdot 3^{-2x} = \frac{2}{3^{2x}}$$ 7. **Rewrite the expression:** $$\frac{1}{2} - \frac{2}{3^x} + \frac{2}{3^{2x}}$$ 8. **Evaluate the limit as $x \to \infty$:** Since $3^x \to \infty$, both $\frac{2}{3^x} \to 0$ and $\frac{2}{3^{2x}} \to 0$. Therefore, $$\lim_{x \to \infty} \frac{(3^x - 2)^2}{2 \cdot 9^x} = \frac{1}{2} - 0 + 0 = \frac{1}{2}$$ **Final answer:** $$\boxed{\frac{1}{2}}$$