1. **State the problem:** Find the limit $$\lim_{x \to \infty} \frac{3e^{-x} + 4e^{x}}{4e^{-x} + 2e^{x}}.$$\n\n2. **Recall the behavior of exponential functions:** As $x \to \infty$, $e^{x} \to \infty$ and $e^{-x} = \frac{1}{e^{x}} \to 0$.\n\n3. **Rewrite the expression to simplify:** Divide numerator and denominator by $e^{x}$ (the dominant term) to handle the limit:\n$$\frac{3e^{-x} + 4e^{x}}{4e^{-x} + 2e^{x}} = \frac{\frac{3e^{-x}}{e^{x}} + \frac{4e^{x}}{e^{x}}}{\frac{4e^{-x}}{e^{x}} + \frac{2e^{x}}{e^{x}}} = \frac{3e^{-2x} + 4}{4e^{-2x} + 2}.$$\n\n4. **Simplify the expression:**\n$$\frac{3e^{-2x} + 4}{4e^{-2x} + 2}.$$\n\n5. **Evaluate the limit as $x \to \infty$:** Since $e^{-2x} \to 0$, the expression becomes\n$$\frac{3 \cdot 0 + 4}{4 \cdot 0 + 2} = \frac{4}{2} = 2.$$\n\n**Final answer:** $$\boxed{2}.$$