Question

use the graph of f(x) to find the limit.
\lim_{x \to 8} f(x) = \square

Explanation:

Step1: Recall Limit Definition

The limit of a function as \( x \) approaches a value (here \( x \to 8 \)) depends on the behavior of the function near \( x = 8 \), not the value at \( x = 8 \).

Step2: Analyze Graph Near \( x = 8 \)

Looking at the graph, as \( x \) approaches 8 from both left and right, the function approaches the open circle (which represents the limit - the closed dot is the function's value at \( x = 8 \), but limit ignores that). The open circle at \( x = 8 \) is at \( y = 5 \) (since the line has a slope, and from the grid, at \( x = 8 \), the limit - the approaching value - is 5). Wait, let's check the graph again. Wait, the line: when \( x = 0 \), \( y = 1 \) (wait, no, the y-intercept: looking at the graph, the line crosses the y-axis at \( y = 1 \)? Wait, no, let's calculate the slope. Wait, from \( x = -4 \), \( y = 0 \), and \( x = 0 \), \( y = 1 \)? Wait, no, maybe my initial reading was wrong. Wait, the graph: the line goes from the left, crosses the x-axis at \( x = -4 \), then goes up. At \( x = 8 \), the open circle is at \( y = 5 \)? Wait, let's see the grid. Each square is 1 unit. So from \( x = 0 \), \( y = 1 \) (wait, no, when \( x = 0 \), the y-coordinate is 1? Wait, no, the line passes through (0,1)? Wait, no, looking at the graph, when \( x = 8 \), the open circle is at \( y = 5 \)? Wait, no, let's check the coordinates. The line: let's take two points. When \( x = -4 \), \( y = 0 \); when \( x = 0 \), \( y = 1 \)? Wait, no, maybe the y-intercept is 1? Wait, no, the graph shows that at \( x = 8 \), the open circle is at \( y = 5 \)? Wait, no, let's count the grid. From \( x = 0 \) to \( x = 8 \), the x increases by 8, and the y increases by 4? Wait, no, the slope: from \( x = -4 \) (y=0) to \( x = 4 \) (y=2), slope is \( \frac{2 - 0}{4 - (-4)} = \frac{2}{8} = \frac{1}{4} \). Then at \( x = 8 \), \( y = 0 + \frac{1}{4}(8 - (-4)) = \frac{1}{4}(12) = 3 \)? Wait, no, maybe I messed up. Wait, the graph: the line goes through ( -4, 0 ) and (8, 5)? No, the open circle at \( x = 8 \) is at \( y = 5 \)? Wait, looking at the graph, the open circle is at (8,5) and the closed dot is at (8,2). Wait, no, the closed dot is at (8,2), open circle at (8,5). So as \( x \) approaches 8 from left and right, the function approaches 5 (the open circle), so the limit is 5? Wait, no, wait, let's re-express. The limit as \( x \to 8 \) is the value the function approaches, regardless of the value at \( x = 8 \). So the graph near \( x = 8 \): from the left, the line is approaching the open circle at (8,5), and from the right, also approaching (8,5). So the limit is 5? Wait, no, maybe my grid reading is wrong. Wait, the y-axis: each grid line is 1 unit. So at \( x = 8 \), the open circle is at \( y = 5 \) (since from \( x = 0 \), \( y = 1 \), then slope \( \frac{5 - 1}{8 - 0} = \frac{4}{8} = 0.5 \). So the equation is \( y = 0.5x + 1 \). Then at \( x = 8 \), \( y = 0.5*8 + 1 = 4 + 1 = 5 \). Yes, that makes sense. So the limit as \( x \to 8 \) is 5, because the function approaches 5 as \( x \) gets close to 8 from both sides.

Answer:

5