1. **State the problem:** We need to find the limit as $n$ approaches infinity of the expression $$\frac{1 + (-1)^n}{2 + (-1)^n}.$$\n\n2. **Analyze the behavior of $(-1)^n$:** The term $(-1)^n$ alternates between $1$ and $-1$ depending on whether $n$ is even or odd.\n- If $n$ is even, $(-1)^n = 1$.\n- If $n$ is odd, $(-1)^n = -1$.\n\n3. **Evaluate the expression for even $n$: $$\frac{1 + 1}{2 + 1} = \frac{2}{3}.$$**\n\n4. **Evaluate the expression for odd $n$: $$\frac{1 + (-1)}{2 + (-1)} = \frac{0}{1} = 0.$$**\n\n5. **Determine the limit:** Since the expression oscillates between $\frac{2}{3}$ (for even $n$) and $0$ (for odd $n$), the limit does not approach a single value as $n$ goes to infinity.\n\n**Final answer:** The limit $$\lim_{n \to \infty} \frac{1 + (-1)^n}{2 + (-1)^n}$$ does not exist because the sequence oscillates between two values.