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Let's analyze and solve each part of the task based on the given function \( f(x) \) defined piecewise as follows:

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

### Part (a) 
**Find the limit as \( x \) approaches \( 0 \) from the left (i.e., \( \lim_{x \to 0^-} f(x) \)).**

Since for \( x < 0 \), \( f(x) = 1 \), the limit from the left as \( x \) approaches \( 0 \) is:
\[
\lim_{x \to 0^-} f(x) = 1
\]

### Part (b)
**Find the limit as \( x \) approaches \( 0 \) from the right (i.e., \( \lim_{x \to 0^+} f(x) \)).**

For \( x \) in the interval \( 0 \leq x < 3 \), \( f(x) = x^2 - 3x \). Evaluating the limit as \( x \) approaches \( 0 \) from the right:
\[
\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x^2 - 3x) = 0^2 - 3 \times 0 = 0
\]

### Part (c)
**Find the limit as \( x \) approaches \( 2 \) (i.e., \( \lim_{x \to 2} f(x) \)).**

Since \( 2 \) is within the interval \( 0 \leq x < 3 \) and the function is continuous there:
\[
\lim_{x \to 2} f(x) = 2^2 - 3 \times 2 = 4 - 6 = -2
\]

### Part (d)
**Determine the type of discontinuity, if any, at \( x = 3 \).**

To analyze the continuity at \( x = 3 \), check the left-hand and right-hand limits:
\[
\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (x^2 - 3x) = 3^2 - 3 \times 3 = 9 - 9 = 0
\]
\[
\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (2x) = 2 \times 3 = 6
\]

Since \( \lim_{x \to 3^-} f(x) \neq \lim_{x \to 3^+} f(x) \), there is a discontinuity at \( x = 3 \), and it is a **jump discontinuity**.

**Summary of Answers:**
- \( \lim_{x \to 0^-} f(x) = 1 \)
- \( \lim_{x \to 0^+} f(x) = 0 \)
- \( \lim_{x \to 2} f(x) = -2 \)
- At \( x = 3 \), there is a jump discontinuity.

If you have any questions about these calculations or need further details, feel free to ask! Here are some follow-up topics you might find interesting:
1. What would the graph of this function look like?
2. How does the nature of the discontinuity affect the integrability of the function?
3. Can you provide an example of an infinite discontinuity and a removable discontinuity?
4. How can limits help in understanding the behavior of functions not just at points, but also over intervals?
5. Why is the limit concept fundamental in calculus?

**Tip:** When analyzing a piecewise function, always check the behavior near the boundaries of the definition intervals to understand limits and continuity.

Consider the following function: f(x) = {1/x, if x < 0; x^2 - 3x, if 0 ≤ x < 3; 2x, if x ≥ 3}. Find the following limits: a) lim x->0- f(x), b) lim x->0+ f(x), c) lim x->2 f(x). Determine the type of discontinuity at x = 3.

Explore how to calculate the limits of a piecewise function and identify the type of discontinuity at specific points. This problem involves limits, continuity, and the analysis of piecewise-defined functions, suitable for advanced high school students.

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