1. **Problem Statement:** Test the continuity of the piecewise function $$f(x) = \begin{cases} \frac{x^2 - 4}{x - 2} & x < 2 \\ \frac{5}{3}x + \frac{2}{3} & 2 \leq x < 5 \\ x^2 - 16 & x \geq 5 \end{cases}$$ at points $x=2$ and $x=5$. 2. **Recall continuity conditions:** A function $f$ is continuous at $x=c$ if: - $f(c)$ is defined. - $\lim_{x \to c} f(x)$ exists. - $\lim_{x \to c} f(x) = f(c)$. 3. **Check continuity at $x=2$: ** - Calculate left-hand limit (LHL) as $x \to 2^-$: $$\lim_{x \to 2^-} \frac{x^2 - 4}{x - 2} = \lim_{x \to 2^-} \frac{(x-2)(x+2)}{x-2}$$ Cancel common factor: $$\lim_{x \to 2^-} \frac{\cancel{(x-2)}(x+2)}{\cancel{(x-2)}} = \lim_{x \to 2^-} (x+2) = 4$$ - Calculate right-hand limit (RHL) as $x \to 2^+$: $$\lim_{x \to 2^+} \left( \frac{5}{3}x + \frac{2}{3} \right) = \frac{5}{3} \times 2 + \frac{2}{3} = \frac{10}{3} + \frac{2}{3} = 4$$ - Calculate $f(2)$: Since $2 \leq x < 5$, use second piece: $$f(2) = \frac{5}{3} \times 2 + \frac{2}{3} = 4$$ - Since LHL = RHL = $f(2) = 4$, $f$ is continuous at $x=2$. 4. **Check continuity at $x=5$: ** - Calculate left-hand limit (LHL) as $x \to 5^-$: $$\lim_{x \to 5^-} \left( \frac{5}{3}x + \frac{2}{3} \right) = \frac{5}{3} \times 5 + \frac{2}{3} = \frac{25}{3} + \frac{2}{3} = 9$$ - Calculate right-hand limit (RHL) as $x \to 5^+$: $$\lim_{x \to 5^+} (x^2 - 16) = 5^2 - 16 = 25 - 16 = 9$$ - Calculate $f(5)$: Since $x \geq 5$, use third piece: $$f(5) = 5^2 - 16 = 9$$ - Since LHL = RHL = $f(5) = 9$, $f$ is continuous at $x=5$. **Final answer:** The function $f$ is continuous at both $x=2$ and $x=5$.