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### Problem c)
The problem now involves a piecewise function \( f(x) \) defined as:

\[
f(x) = 
\begin{cases} 
\sqrt{x} & \text{if } x < 0, \\
3 - x & \text{if } 0 \leq x < 3, \\
3x - 2 & \text{if } x \geq 3.
\end{cases}
\]

The subproblems are:

1. **Find \(\lim_{x \to 3} f(x)\) if it exists.**

2. **Show that \( f(x) \) is continuous at \( x = 3 \).**

3. **Sketch the graph of \( f(x) \).**

#### 1. Find \(\lim_{x \to 3} f(x)\)
To determine whether \(\lim_{x \to 3} f(x)\) exists, we need to check the left-hand limit (\(\lim_{x \to 3^-} f(x)\)) and the right-hand limit (\(\lim_{x \to 3^+} f(x)\)).

- **Left-hand limit (\(x \to 3^-\))**: In this region, \(f(x) = 3 - x\). So,
  \[
  \lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (3 - x) = 3 - 3 = 0.
  \]

- **Right-hand limit (\(x \to 3^+\))**: In this region, \(f(x) = 3x - 2\). So,
  \[
  \lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (3x - 2) = 3(3) - 2 = 9 - 2 = 7.
  \]

Since the left-hand limit (0) and right-hand limit (7) are not equal, \(\lim_{x \to 3} f(x)\) **does not exist**.

#### 2. Show that \( f(x) \) is continuous at \( x = 3 \)
For \( f(x) \) to be continuous at \( x = 3 \), the following must be true:
- \( f(3) \) is defined,
- \(\lim_{x \to 3^-} f(x)\) exists,
- \(\lim_{x \to 3^+} f(x)\) exists,
- \(\lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) = f(3)\).

Let's check these conditions:

- **\(f(3)\) is defined**: From the piecewise function, \( f(3) = 3(3) - 2 = 9 - 2 = 7 \).

- **Left-hand limit**: \(\lim_{x \to 3^-} f(x) = 0\).

- **Right-hand limit**: \(\lim_{x \to 3^+} f(x) = 7\).

Since \(\lim_{x \to 3^-} f(x)\) and \(\lim_{x \to 3^+} f(x)\) are not equal, \( f(x) \) **is not continuous** at \( x = 3 \).

#### 3. Sketch the graph of \( f(x) \)
For sketching the graph, break it down according to the piecewise function:

1. **For \( x < 0 \)**: The function \( f(x) = \sqrt{x} \) is **not defined** for real numbers since the square root of a negative number is not a real number.

2. **For \( 0 \leq x < 3 \)**: The function is \( f(x) = 3 - x \), a linear function with a negative slope of -1. This line decreases as \( x \) increases, starting from \( (0, 3) \) and ending just before \( (3, 0) \).

3. **For \( x \geq 3 \)**: The function is \( f(x) = 3x - 2 \), another linear function but with a positive slope of 3. This line starts at \( (3, 7) \) and continues upward.

**Key Points:**
- There is a **jump discontinuity** at \( x = 3 \).
- The graph is not defined for \( x < 0 \).
- For \( 0 \leq x < 3 \), the graph is a downward sloping line.
- For \( x \geq 3 \), the graph is an upward sloping line starting at \( (3, 7) \).

---

Would you like a more detailed explanation on any of these steps, or perhaps the sketch of the graph? Here are some related questions you might consider:

### Related Questions:
1. What are the conditions necessary for a limit to exist at a point?
2. How can you determine the continuity of a piecewise function?
3. What are the differences between jump discontinuity and removable discontinuity?
4. How does the behavior of limits affect the sketching of a piecewise function?
5. What role does the slope play in determining the direction of a line?

### Tip:
When evaluating limits of piecewise functions, always consider the left-hand and right-hand limits separately before concluding whether the overall limit exists at a particular point.

Explore a detailed step-by-step analysis of a piecewise function involving limits, continuity, and graph sketching. This problem covers key mathematical concepts such as piecewise functions, limits evaluation, and continuity checks. Suitable for advanced high school students (Grades 11-12).

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