1. **State the problem:** Find the limit of the piecewise function $$f(x)$$ as $$x \to 2$$, where $$f(x) = \begin{cases} x - 1, & x < 2 \\ 1, & x = 2 \\ x + 1, & x > 2 \end{cases}$$ 2. **Recall the definition of limit:** The limit $$\lim_{x \to a} f(x)$$ exists if and only if the left-hand limit $$\lim_{x \to a^-} f(x)$$ and the right-hand limit $$\lim_{x \to a^+} f(x)$$ both exist and are equal. 3. **Calculate the left-hand limit:** $$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x - 1) = 2 - 1 = 1$$ 4. **Calculate the right-hand limit:** $$\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (x + 1) = 2 + 1 = 3$$ 5. **Compare the limits:** Since $$\lim_{x \to 2^-} f(x) = 1$$ and $$\lim_{x \to 2^+} f(x) = 3$$, the left and right limits are not equal. 6. **Conclusion:** Because the left and right limits are different, the limit $$\lim_{x \to 2} f(x)$$ does not exist. **Final answer:** The limit does not exist.