1. **State the problem:** Find the following limits and function value from the graph of $g$: - $\lim_{x \to 2^-} g(x)$ - $\lim_{x \to 2^+} g(x)$ - $\lim_{x \to 2} g(x)$ - $\lim_{x \to 0} g(x)$ - $g(2)$ 2. **Recall limit definitions:** - The left-hand limit $\lim_{x \to a^-} g(x)$ is the value $g(x)$ approaches as $x$ approaches $a$ from the left. - The right-hand limit $\lim_{x \to a^+} g(x)$ is the value $g(x)$ approaches as $x$ approaches $a$ from the right. - The limit $\lim_{x \to a} g(x)$ exists only if both left and right limits exist and are equal. - The function value $g(a)$ is the actual value of the function at $x=a$. 3. **Analyze the graph at $x=2$:** - From the left, the graph approaches the open circle at $(2,2)$ but the filled point at $x=2$ is at $(2,1)$. - From the right, the graph starts at an open circle at $(2,0)$ and rises. 4. **Evaluate each limit and value:** - $\lim_{x \to 2^-} g(x) = 2$ (approaches the open circle at $(2,2)$ from the left branch) - $\lim_{x \to 2^+} g(x) = 0$ (approaches the open circle at $(2,0)$ from the right branch) - $\lim_{x \to 2} g(x)$ does not exist because left and right limits differ: $2 \neq 0$ - $\lim_{x \to 0} g(x)$: From the graph, the left branch ends at a filled point $(0,2)$ and the right branch starts at an open circle $(0,-2)$. Since the left limit at $0$ is $2$ and the right limit is $-2$, the limit at $0$ does not exist. - $g(2) = 1$ (the filled point at $x=2$ is at $(2,1)$) **Final answers:** $$\lim_{x \to 2^-} g(x) = 2$$ $$\lim_{x \to 2^+} g(x) = 0$$ $$\lim_{x \to 2} g(x) \text{ does not exist}$$ $$\lim_{x \to 0} g(x) \text{ does not exist}$$ $$g(2) = 1$$