1. **Problem Statement:** We are given a piecewise function: $$f(x) = \begin{cases} |x-2|, & x \neq 2 \\ 4, & x = 2 \end{cases}$$ We want to analyze the function's value at $x=2$, the limit as $x$ approaches 2, and whether the function is continuous at $x=2$. 2. **Recall the definitions:** - $f(a)$ exists means the function has a defined value at $x=a$. - $\lim_{x \to a} f(x)$ exists means the left-hand and right-hand limits at $x=a$ are equal. - The function is continuous at $x=a$ if $f(a)$ exists, $\lim_{x \to a} f(x)$ exists, and $\lim_{x \to a} f(x) = f(a)$. 3. **Evaluate $f(2)$:** Given directly as $f(2) = 4$. 4. **Evaluate the limit as $x \to 2$:** - Left-hand limit: $$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} |x-2| = |2 - 2| = 0$$ - Right-hand limit: $$\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} |x-2| = |2 - 2| = 0$$ Since both limits equal 0, the limit exists and is 0: $$\lim_{x \to 2} f(x) = 0$$ 5. **Check continuity at $x=2$:** - $f(2) = 4$ - $\lim_{x \to 2} f(x) = 0$ Since $f(2) \neq \lim_{x \to 2} f(x)$, the function is **not continuous** at $x=2$. **Final answer:** - $f(2)$ exists and equals 4. - $\lim_{x \to 2} f(x)$ exists and equals 0. - The function is not continuous at $x=2$ because $f(2) \neq \lim_{x \to 2} f(x)$.