1. **State the problem:** We have a piecewise function: $$f(x) = \begin{cases} \frac{(x+2)(x-2)}{x-2}, & x \neq 2 \\ 1, & x = 2 \end{cases}$$ We want to sketch its graph, find $f(2)$, and determine the left-hand and right-hand limits as $x$ approaches 2. 2. **Simplify the function for $x \neq 2$:** Since $x \neq 2$, the denominator $x-2 \neq 0$, so we can simplify: $$f(x) = \frac{(x+2)(x-2)}{x-2} = x + 2$$ 3. **Evaluate $f(2)$:** By definition, $f(2) = 1$. 4. **Find the limits as $x$ approaches 2:** - Left-hand limit ($x \to 2^-$): $$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x + 2) = 2 + 2 = 4$$ - Right-hand limit ($x \to 2^+$): $$\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (x + 2) = 2 + 2 = 4$$ 5. **Interpretation:** The function behaves like the line $y = x + 2$ everywhere except at $x=2$, where there is a hole (removable discontinuity) because the simplified function would give $f(2) = 4$, but the function is defined as $f(2) = 1$. 6. **Graph description:** - The graph is the line $y = x + 2$ with a hole at $(2,4)$. - There is a filled point at $(2,1)$. This matches the problem's description. **Final answers:** $$f(2) = 1$$ $$\lim_{x \to 2^-} f(x) = 4$$ $$\lim_{x \to 2^+} f(x) = 4$$