1. **Problem:** Determine if the function $$f(x) = \begin{cases} x + 2, & x < 1 \\ 1, & x = 1 \\ 2 - x, & x > 1 \end{cases}$$ is continuous at $x = 1$. 2. **Continuity condition:** A function $f$ is continuous at $x = a$ if $$\lim_{x \to a^-} f(x) = f(a) = \lim_{x \to a^+} f(x).$$ 3. **Calculate left-hand limit:** $$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x + 2) = 1 + 2 = 3.$$ 4. **Calculate right-hand limit:** $$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2 - x) = 2 - 1 = 1.$$ 5. **Evaluate function at $x=1$:** $$f(1) = 1.$$ 6. **Check continuity:** Since $$\lim_{x \to 1^-} f(x) = 3 \neq 1 = f(1) = \lim_{x \to 1^+} f(x),$$ the function is **not continuous** at $x = 1$. **Final answer:** The function is not continuous at $x = 1$ because the left-hand limit does not equal the function value or the right-hand limit.