1. The problem asks for the values of $x_0$ in the interval $-3 \leq x_0 \leq 2$ where the limit $\lim_{x \to x_0} g(x)$ exists. 2. The limit $\lim_{x \to x_0} g(x)$ exists if and only if the left-hand limit and right-hand limit at $x_0$ are equal. 3. From the graph description: - At $x_0 = -3$, the function is continuous with a filled dot, so the limit exists. - Between $-3$ and $-1$, the function is a line segment, so the limit exists for all $x_0$ in $(-3, -1)$. - At $x_0 = -1$, there is a jump: left limit approaches 2 (open circle), right limit approaches 1 (filled dot), so the limit does not exist. - Between $-1$ and $0$, the function is a line segment with filled dots at both ends, so the limit exists for $x_0$ in $(-1, 0)$. - At $x_0 = 0$, left limit approaches 0 (filled dot), right limit approaches 0 (open circle), but since the open circle is at 0, the limit exists and equals 0. - Between $0$ and $1$, the function is a line segment with an open circle at 0 and filled dot at 1, so the limit exists for $x_0$ in $(0, 1)$. - At $x_0 = 1$ and $x_0 = 2$, the function has filled dots and continuous segments, so the limit exists. 4. Therefore, the limit exists for all $x_0$ in $[-3, 2]$ except at $x_0 = -1$. Final answer: The limit $\lim_{x \to x_0} g(x)$ exists for all $x_0$ in $[-3, 2]$ except at $x_0 = -1$.