y = f(x) (graph with x - axis 0 - 9, y - axis 0 - 6). enter each answer as a whole number (like - 4, 0, or 253) or dne for undefined or does not exist.
$f(7) = \square$
$lim\limits_{x \to 4} f(x) = \square$
$lim\limits_{x \to 0^+} f(x) = \square$
$lim\limits_{x \to 0} f(x) = \square$
question help: $\boxed{\text{video 1}}$ $\boxed{\text{video 2}}$

Question

Accepted Answer

### 1. Solve $ f(7) $
# Explanation:
## Step1: Find the point at $ x = 7 $
From the graph, at $ x = 7 $, the $ y $-value (function value) is 6.
# Answer:
$ f(7) = 6 $

### 2. Solve $ \lim_{x \to 4} f(x) $
# Explanation:
## Step1: Check left and right limits at $ x = 4 $
As $ x $ approaches 4 from the left, the function approaches 1. As $ x $ approaches 4 from the right, the function approaches 3. Since left and right limits differ, the limit does not exist (DHE).
# Answer:
$ \lim_{x \to 4} f(x) = \text{DHE} $

### 3. Solve $ \lim_{x \to 0^+} f(x) $
# Explanation:
## Step1: Analyze the right - hand limit at $ x = 0 $
As $ x $ approaches 0 from the right, we look at the behavior of the function. The function starts from a point near $ (0, 2.5) $ but as $ x $ approaches 0 from the right (moving along the graph from the right of $ x = 0 $ towards $ x = 0 $), the $ y $-value approaches 3? Wait, no, re - examining the graph: the leftmost part of the graph starts at a point with $ y\approx2.5 $ but when approaching $ x = 0 $ from the right (i.e., $ x>0 $ getting closer to 0), the first segment is from $ x = 0 $ (right - hand side) to $ x = 1 $, with the point at $ x = 1 $ being $ (1,1) $ and the left - hand side of the first segment (near $ x = 0 $) has a $ y $-value approaching 3? Wait, no, the first part of the graph: from $ x = 0 $ (the left - hand side of the graph starts at $ y\approx2.5 $) but when we take $ x\to0^+ $, we look at the function values for $ x $ just greater than 0. The first segment (from $ x = 0 $ to $ x = 1 $): at $ x = 1 $, $ f(1)=1 $, and as $ x $ approaches 0 from the right, the $ y $-value approaches 3? Wait, no, the graph has a segment from $ (0, 2.5) $ (approx) going up to $ (1,4) $? Wait, no, the graph: at $ x = 1 $, there are two points? Wait, no, the red graph: let's re - interpret. The first part: from $ x = 0 $ (left - hand side, $ y\approx2.5 $) to $ x = 1 $, the point at $ x = 1 $ is $ (1,4) $? Wait, no, the user's graph: the first red segment goes from a point (left of $ x = 1 $) up to $ (1,4) $, and another segment from $ (1,1) $ up to $ (3,3) $. Wait, maybe I misread. Let's start over.

Looking at the graph:

- For $ f(7) $: at $ x = 7 $, the red dot is at $ y = 6 $, so $ f(7)=6 $.

- For $ \lim_{x\to4}f(x) $: as $ x $ approaches 4 from the left, the function value approaches 1 (the dot at $ x = 4 $ left - hand side is 1), and from the right, it approaches 3 (the dot at $ x = 4 $ right - hand side is 3). So left limit $ LHL = 1 $, right limit $ RHL=3 $, $ LHL
eq RHL $, so limit DHE.

- For $ \lim_{x\to0^+}f(x) $: as $ x $ approaches 0 from the right (i.e., $ x>0 $ and $ x $ gets closer to 0), we look at the function values. The first segment of the graph (for $ x $ near 0 from the right) is the one that starts from $ x = 0 $ (right - hand side) and goes up. The $ y $-value at $ x = 0 $ (right - hand) is approaching 3? Wait, the left - most part of the graph (before $ x = 1 $): the first red line goes from a point (let's say $ x = 0 $ - left, $ y\approx2.5 $) up to $ (1,4) $, and another red line from $ (1,1) $ up to $ (3,3) $. Wait, maybe the graph has two parts at $ x = 1 $: one point at $ (1,4) $ and one at $ (1,1) $. So when $ x\to0^+ $, we are looking at the function values for $ x $ just greater than 0, which is the segment that goes from $ x = 0 $ (right) to $ x = 1 $ (the upper segment with $ f(1) = 4 $). So as $ x\to0^+ $, the function approaches 3 (since at $ x = 0 $, the left - hand side of the graph is at $ y\approx2.5 $, but the right - hand side (as $ x$ comes from the right of 0) is part of the segment that starts at $ x = 0 $ (right) with $ y$ approaching 3? Wait, maybe the initial point of the first segment (the one going to $ (1,4) $) is at $ x = 0 $ with $ y = 3 $ (since the graph starts at $ y = 3 $ on the left of $ x = 0 $, and then the first segment (for $ x\geq0 $?) goes up to $ (1,4) $). So as $ x\to0^+ $, the function approaches 3.

- For $ \lim_{x\to0}f(x) $: we need to check left and right limits. The left limit (as $ x\to0^- $): the graph approaches $ y = 3 $ (since the left - most part of the graph is at $ y\approx3 $ when $ x$ approaches 0 from the left). The right limit (as $ x\to0^+ $): as we saw, approaches 3. So $ \lim_{x\to0}f(x)=3 $.

### 3. Solve $ \lim_{x \to 0^+} f(x) $
# Explanation:
## Step1: Analyze right - hand limit at $ x = 0 $
As $ x $ approaches 0 from the right, the function values approach 3 (from the graph's right - hand side of $ x = 0 $ segment).
# Answer:
$ \lim_{x \to 0^+} f(x) = 3 $

### 4. Solve $ \lim_{x \to 0} f(x) $
# Explanation:
## Step1: Check left and right limits at $ x = 0 $
Left limit ( $ x\to0^- $): function approaches 3. Right limit ( $ x\to0^+ $): function approaches 3. Since $ LHL = RHL=3 $, the limit is 3.
# Answer:
$ \lim_{x \to 0} f(x) = 3 $

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QUESTION IMAGE

Question

Explanation:

1. Solve \( f(7) \)

Step1: Find the point at \( x = 7 \)

Step1: Check left and right limits at \( x = 4 \)

Step1: Analyze the right - hand limit at \( x = 0 \)

Answer:

2. Solve \( \lim_{x \to 4} f(x) \)