1. **State the problem:** We are given a graph of a function $f(x)$ and need to find the left-hand limit, right-hand limit, two-sided limit at $x = -3$, and determine where $f$ is continuous. 2. **Recall limit definitions:** - The left-hand limit $\lim_{x \to a^-} f(x)$ is the value $f(x)$ approaches as $x$ approaches $a$ from the left. - The right-hand limit $\lim_{x \to a^+} f(x)$ is the value $f(x)$ approaches as $x$ approaches $a$ from the right. - The two-sided limit $\lim_{x \to a} f(x)$ exists if and only if the left and right limits are equal. - A function is continuous at $x=a$ if $\lim_{x \to a} f(x) = f(a)$. 3. **Analyze the graph at $x = -3$:** - From the left ($x \to -3^-$), the curve approaches the filled point near $(-3, 1.5)$. - From the right ($x \to -3^+$), the curve approaches the open circle near $(-3, 2.5)$. - The filled point at $(-3, 1.5)$ means $f(-3) = 1.5$. 4. **Evaluate limits:** - $\lim_{x \to -3^-} f(x) = 1.5$ - $\lim_{x \to -3^+} f(x) = 2.5$ 5. **Check two-sided limit:** Since $1.5 \neq 2.5$, $$\lim_{x \to -3} f(x) \text{ does not exist}$$ 6. **Determine continuity at $x = -3$:** - Because the two-sided limit does not exist, $f$ is not continuous at $x = -3$. 7. **Where is $f$ continuous?** - The graph is smooth and connected except at $x = -3$ where there is a jump. - Therefore, $f$ is continuous everywhere except at $x = -3$. **Final answers:** (a) $\lim_{x \to -3^-} f(x) = 1.5$ (b) $\lim_{x \to -3^+} f(x) = 2.5$ (c) $\lim_{x \to -3} f(x)$ does not exist (d) $f$ is continuous for all $x \neq -3$