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The problem in the image is about finding the limit of a function using factoring techniques. Here's a detailed breakdown of the process shown:

### Problem:
Determine the limit:
\[
\lim_{x \to 1} \frac{3x - 3}{x^2 - 4x + 3}
\]

### Solution:

1. **Substitution of \( x = 1 \) (Direct Substitution):**
   \[
   \frac{3(1) - 3}{1^2 - 4(1) + 3} = \frac{3 - 3}{1 - 4 + 3} = \frac{0}{0}
   \]
   Since the result is \( \frac{0}{0} \) (indeterminate form), we need to simplify the expression by factoring.

2. **Factoring the numerator and denominator:**
   - Numerator: \( 3x - 3 = 3(x - 1) \)
   - Denominator: \( x^2 - 4x + 3 \) factors as \( (x - 1)(x - 3) \)
   
   Now the limit becomes:
   \[
   \lim_{x \to 1} \frac{3(x - 1)}{(x - 1)(x - 3)}
   \]
   
3. **Cancel out the common factor \( (x - 1) \):**
   After canceling \( (x - 1) \) from both the numerator and denominator:
   \[
   \lim_{x \to 1} \frac{3}{x - 3}
   \]

4. **Substitute \( x = 1 \) into the simplified expression:**
   \[
   \frac{3}{1 - 3} = \frac{3}{-2} = -\frac{3}{2}
   \]

### Final Answer:
The limit is:
\[
\lim_{x \to 1} \frac{3x - 3}{x^2 - 4x + 3} = -\frac{3}{2}
\]

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Would you like further details or any clarifications on this solution?

Here are 5 follow-up questions to deepen your understanding:

1. What is the indeterminate form, and why does it require additional steps to solve?
2. How do you recognize when factoring can be used to simplify a limit problem?
3. Can you always cancel out terms like \( (x - 1) \) in the numerator and denominator?
4. What other techniques, besides factoring, can help resolve limits involving indeterminate forms?
5. What happens if the limit cannot be resolved by factoring alone?

**Tip**: When faced with an indeterminate form, always try factoring or using algebraic manipulation to simplify the expression before considering more advanced methods like L'Hopital's Rule.

Menentukan nilai limit: lim_{x → 1} (3x - 3) / (x^2 - 4x + 3)

Learn how to determine the limit of (3x - 3) / (x^2 - 4x + 3) as x approaches 1. This detailed solution involves factoring both the numerator and the denominator to resolve the indeterminate form 0/0. Suitable for students in Grades 10-12.

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