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: $\lim_{x \to 1} \frac{3x - 3}{x^2 - 4x + 3} = -\frac{3}{2}$

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.