Question

1. consider the function g graphed in the accompanying figure. for what values of ( x_0 ) does ( lim_{x \to x_0} g(x) ) exist? figure ex-13

Explanation:

Step1: Recall Limit Existence Rule

The limit $\lim_{x \to x_0} g(x)$ exists if and only if the left - hand limit $\lim_{x \to x_0^-} g(x)$ and the right - hand limit $\lim_{x \to x_0^+} g(x)$ both exist and are equal, i.e., $\lim_{x \to x_0^-} g(x)=\lim_{x \to x_0^+} g(x)$.

Step2: Analyze the Graph at $x_0 = - 4$

• For $x_0=-4$:
• The left - hand limit: As $x$ approaches $-4$ from the left, we follow the left - hand part of the graph. The left - hand side of $x = - 4$ is a line segment. The left - hand limit $\lim_{x \to - 4^-}g(x)$ is the $y$ - value that the left - hand part approaches.
• The right - hand limit: As $x$ approaches $-4$ from the right, we follow the right - hand part of the graph (the downward - sloping line starting from the open circle at $x=-4$). The left - hand limit (from the left of $x = - 4$) and the right - hand limit (from the right of $x=-4$) are not equal. Because the left - hand side approaches the $y$ - value of the closed dot at $x=-4$, and the right - hand side approaches the $y$ - value of the open dot at $x = - 4$, which are different. So, the limit does not exist at $x_0=-4$.

Step3: Analyze the Graph at $x_0 = 2$

• For $x_0 = 2$:
• The left - hand limit: As $x$ approaches $2$ from the left, we follow the part of the graph (the line from the origin to the open circle at $x = 2$). The left - hand limit $\lim_{x \to 2^-}g(x)$ is the $y$ - value of the open circle at $x = 2$ (which is $2$).
• The right - hand limit: As $x$ approaches $2$ from the right, we follow the right - hand part of the graph (the line going upwards from $x = 2$). The right - hand limit $\lim_{x \to 2^+}g(x)$ is equal to the left - hand limit (because the line from the origin to the open circle at $x = 2$ and the line from $x = 2$ onwards have the same slope and the left - hand limit (value of the open circle) and the right - hand limit (value approached from the right) are equal). Wait, actually, looking at the graph: the left - hand side (as $x$ approaches $2$ from the left) follows the line that comes from the origin and reaches the open circle at $x = 2$ (with $y = 2$), and the right - hand side (as $x$ approaches $2$ from the right) follows the line that goes upwards. The left - hand limit and the right - hand limit are equal (both approach the $y$ - value of the open circle at $x = 2$), even though the function value at $x = 2$ (the closed dot) is different. So the limit exists at $x_0=2$.

Step4: Analyze Other Points

• For all other values of $x_0$ (i.e., $x_0

eq - 4$):

• The graph of $g(x)$ is composed of line segments (except at $x=-4$ and $x = 2$, but we saw $x = 2$ is okay). For any $x_0$ not equal to $-4$, the left - hand limit and the right - hand limit will be equal. Because at points other than $x=-4$, the function is either a single line segment (so left and right limits are the same as the function is continuous there in terms of the limit - the function may have a different value at a point, but the limit depends on the approach) or the two adjacent line segments (like around $x = 0$) have the same left and right limits (since the graph meets at $x = 0$ with the same slope from both sides? Wait, at $x = 0$, the graph has a "corner" but the left - hand limit (approaching from the left, along the line from $x=-4$ to $x = 0$) and the right - hand limit (approaching from the right, along the line from $x = 0$ to $x = 2$) are equal (both approach $0$). In general, the only point where the left - hand limit and the right - hand limit differ is at $x=-4$.

Answer:

The limit $\lim_{x \to x_0}g(x)$ exists for all real numbers $x_0$ except $x_0=-4$, i.e., $x_0\in\mathbb{R}\setminus\{-4\}$ or $x_0
eq - 4$.