1. **State the problem:** We need to determine if the piecewise function $$h(x) = \begin{cases} x - 4 & \text{if } x < 2 \\ x^2 - 16 & \text{if } x \geq 2 \end{cases}$$ is continuous at $x = -4$. 2. **Calculate $h(-4)$:** Since $-4 < 2$, use the first piece: $$h(-4) = (-4) - 4 = -8$$ 3. **Find the limit $\lim_{x \to -4} h(x)$:** Since $-4 < 2$, for values near $-4$, we use the first piece $x - 4$. Calculate the limit from the left and right: - Left limit: $\lim_{x \to -4^-} h(x) = \lim_{x \to -4^-} (x - 4) = -4 - 4 = -8$ - Right limit: $\lim_{x \to -4^+} h(x) = \lim_{x \to -4^+} (x - 4) = -4 - 4 = -8$ Since both one-sided limits are equal, the limit exists: $$\lim_{x \to -4} h(x) = -8$$ 4. **Check the three-part test for continuity at $x = -4$:** - (1) $h(-4)$ is defined and equals $-8$. - (2) $\lim_{x \to -4} h(x)$ exists and equals $-8$. - (3) $h(-4) = \lim_{x \to -4} h(x) = -8$. 5. **Conclusion:** The function $h(x)$ is continuous at $x = -4$ because it passes all three conditions for continuity.