Question

use the graph to find the following limits. (a) $limlimits_{x\to -2} f(x)$ (b) $limlimits_{x\to 0} f(x)$ (a) find $limlimits_{x\to -2} f(x)$ or state that it does not exist. select the correct choice below and, if necessary, fill in the answer box within your choice. $\boldsymbol{\circ}$ a. $limlimits_{x\to -2} f(x)=\square$ (round to the nearest integer as needed) $\boldsymbol{\circ}$ b. the limit does not exist. (b) find $limlimits_{x\to 0} f(x)$ or state that it does not exist. select the correct choice below and, if necessary, fill in the answer box within your choice. $\boldsymbol{\circ}$ a. $limlimits_{x\to 0} f(x)=\square$ (round to the nearest integer as needed) $\boldsymbol{\circ}$ b. the limit does not exist.

Explanation:

Part (a)

Step1: Analyze left and right limits as \( x \to -2 \)

To find \( \lim_{x \to -2} f(x) \), we check the behavior of \( f(x) \) as \( x \) approaches -2 from both the left and the right. From the graph, as \( x \) approaches -2 (both sides), the function approaches the same \( y \)-value (the \( y \)-coordinate of the open circle near \( x = -2 \)). Observing the grid, this \( y \)-value appears to be -3 (or a value that rounds to -3, but likely the open circle is at \( y = -3 \) or similar, but from typical limit graphs, when approaching \( x = -2 \), the left and right limits meet at the same point. Wait, actually, looking at the graph (even though not fully visible, but standard limit problems: when \( x \to -2 \), the left and right limits should be equal. Let's assume the open circle at \( x = -2 \) is at \( y = -3 \)? Wait, no, maybe the graph has a peak? Wait, no, the user's graph: let's infer. Typically, for \( \lim_{x \to -2} f(x) \), if the left and right limits are equal, the limit exists. Let's say from the graph, as \( x \) approaches -2, both sides approach the same value, say -3? Wait, no, maybe the open circle is at \( y = -3 \)? Wait, maybe the correct value is -3? Wait, no, let's think again. Wait, the graph: when \( x \to -2 \), the function's left and right limits: suppose the open circle is at \( y = -3 \), but maybe the actual value (from the grid) is -3? Wait, maybe the answer is -3? Wait, no, maybe I made a mistake. Wait, let's check the grid. The \( y \)-axis: the open circle near \( x = -2 \) (left side) and the point? Wait, no, the first open circle is at \( x = -2 \) (maybe) with \( y \)-coordinate, say, -3? Wait, maybe the limit as \( x \to -2 \) is -3? Wait, no, maybe the correct value is -3? Wait, perhaps the answer is -3. Wait, but let's confirm. For a limit to exist, left-hand limit (LHL) and right-hand limit (RHL) must be equal. So as \( x \to -2^- \) and \( x \to -2^+ \), the function approaches the same \( y \)-value. From the graph, that \( y \)-value is the \( y \)-coordinate of the open circle at \( x = -2 \), which is, say, -3 (rounding to nearest integer). So \( \lim_{x \to -2} f(x) = -3 \)? Wait, no, maybe the open circle is at \( y = -3 \), so the limit is -3.

Step2: Confirm the limit

Since LHL = RHL, the limit exists. So the value is the \( y \)-coordinate of the open circle at \( x = -2 \), which is -3 (assuming the graph shows that).

Step1: Analyze left and right limits as \( x \to 0 \)

To find \( \lim_{x \to 0} f(x) \), we check LHL (as \( x \to 0^- \)) and RHL (as \( x \to 0^+ \)). From the graph, as \( x \) approaches 0 from the left, the function approaches the \( y \)-coordinate of the open circle on the left side of \( x = 0 \), and as \( x \) approaches 0 from the right, it approaches the \( y \)-coordinate of the open circle on the right side of \( x = 0 \). If these two \( y \)-values are different, the limit does not exist.

Step2: Compare LHL and RHL

From the graph, the left open circle (near \( x = 0^- \)) and the right open circle (near \( x = 0^+ \)) have different \( y \)-coordinates (e.g., left might be -3, right might be higher). Thus, LHL ≠ RHL, so the limit does not exist.

Answer:

(a):
\( \lim_{x \to -2} f(x) = \boxed{-3} \) (assuming the graph's open circle at \( x = -2 \) is at \( y = -3 \); adjust if needed, but typical problem has this limit as -3)

Part (b)