1. **State the problem:** We want to find the limit as $x \to \infty$ of $$\left(\frac{4^x + 9^x}{x}\right)^{\frac{1}{x}}.$$\n\n2. **Analyze the expression inside the parentheses:** For large $x$, between $4^x$ and $9^x$, the term $9^x$ dominates because $9 > 4$. So, $$4^x + 9^x \approx 9^x.$$\n\n3. **Rewrite the expression using this approximation:** $$\left(\frac{4^x + 9^x}{x}\right)^{\frac{1}{x}} \approx \left(\frac{9^x}{x}\right)^{\frac{1}{x}} = \left(9^x \cdot \frac{1}{x}\right)^{\frac{1}{x}} = \left(9^x\right)^{\frac{1}{x}} \cdot \left(\frac{1}{x}\right)^{\frac{1}{x}}.$$\n\n4. **Simplify each part:** $$\left(9^x\right)^{\frac{1}{x}} = 9^{x \cdot \frac{1}{x}} = 9,$$ and $$\left(\frac{1}{x}\right)^{\frac{1}{x}} = e^{\frac{1}{x} \ln \frac{1}{x}} = e^{-\frac{\ln x}{x}}.$$\n\n5. **Evaluate the limit of the second factor:** As $x \to \infty$, $$-\frac{\ln x}{x} \to 0,$$ so $$e^{-\frac{\ln x}{x}} \to e^0 = 1.$$\n\n6. **Combine the results:** $$\lim_{x \to \infty} \left(\frac{4^x + 9^x}{x}\right)^{\frac{1}{x}} = 9 \times 1 = 9.$$\n\n**Final answer:** $$\boxed{9}.$$