1. **Problem Statement:** Given the piecewise function: $$f(x) = \begin{cases} |x-2|, & x \neq 2 \\ 4, & x = 2 \end{cases}$$ We want to analyze: (i) Whether $f(a)$ exists. (ii) Whether $\lim_{x \to a} f(x)$ exists. (iii) Whether $\lim_{x \to a} f(x) = f(a)$. 2. **Step (i): Check if $f(2)$ exists.** By definition, $f(2) = 4$. So, $f(2)$ exists and equals 4. 3. **Step (ii): Find $\lim_{x \to 2} f(x)$.** Since $f(x) = |x-2|$ for $x \neq 2$, we evaluate the left and right limits: - Left limit: $$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} |x-2| = |2 - 2| = 0$$ - Right limit: $$\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} |x-2| = |2 - 2| = 0$$ Since both one-sided limits equal 0, the limit exists and is 0: $$\lim_{x \to 2} f(x) = 0$$ 4. **Step (iii): Check if $\lim_{x \to 2} f(x) = f(2)$.** We have: $$\lim_{x \to 2} f(x) = 0 \neq f(2) = 4$$ Therefore, the limit of $f(x)$ as $x$ approaches 2 does not equal $f(2)$. 5. **Summary:** - $f(2)$ exists and equals 4. - $\lim_{x \to 2} f(x)$ exists and equals 0. - $\lim_{x \to 2} f(x) \neq f(2)$, so the function is not continuous at $x=2$. This explains the behavior of the function and its limits at $x=2$.