Let's go through the solutions for the limits presented in the image:

1. $\lim_{x \to 1} \frac{x^2 - 1}{x - 1}$

Factor the numerator: $x^2 - 1 = (x - 1)(x + 1)$ So, the limit becomes: $\lim_{x \to 1} \frac{(x - 1)(x + 1)}{x - 1}$ The $x - 1$ terms cancel out, leaving: $\lim_{x \to 1} (x + 1) = 2$

2. $\lim_{x \to 3} \frac{x^2 - 4x + 3}{x^2 - 9}$

Factor both the numerator and denominator: $x^2 - 4x + 3 = (x - 1)(x - 3) \quad \text{and} \quad x^2 - 9 = (x - 3)(x + 3)$ Now the limit becomes: $\lim_{x \to 3} \frac{(x - 1)(x - 3)}{(x - 3)(x + 3)}$ Cancel the $x - 3$ terms: $\lim_{x \to 3} \frac{x - 1}{x + 3} = \frac{3 - 1}{3 + 3} = \frac{2}{6} = \frac{1}{3}$

3. $\lim_{x \to -2} \frac{x^2 + 6x + 8}{x^3 + 8}$

Factor the numerator and denominator: $x^2 + 6x + 8 = (x + 2)(x + 4) \quad \text{and} \quad x^3 + 8 = (x + 2)(x^2 - 2x + 4)$ Now the limit becomes: $\lim_{x \to -2} \frac{(x + 2)(x + 4)}{(x + 2)(x^2 - 2x + 4)}$ Cancel the $x + 2$ terms: $\lim_{x \to -2} \frac{x + 4}{x^2 - 2x + 4}$ Substitute $x = -2$: $\frac{-2 + 4}{(-2)^2 - 2(-2) + 4} = \frac{2}{4 + 4 + 4} = \frac{2}{12} = \frac{1}{6}$

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

Factor both the numerator and denominator: $x^2 - 1 = (x - 1)(x + 1) \quad \text{and} \quad x^3 + 1 = (x + 1)(x^2 - x + 1)$ Now the limit becomes: $\lim_{x \to -1} \frac{(x - 1)(x + 1)}{(x + 1)(x^2 - x + 1)}$ Cancel the $x + 1$ terms: $\lim_{x \to -1} \frac{x - 1}{x^2 - x + 1}$ Substitute $x = -1$: $\frac{-1 - 1}{(-1)^2 - (-1) + 1} = \frac{-2}{1 + 1 + 1} = \frac{-2}{3}$

5. $\lim_{x \to 3} \frac{2x - 6}{x^2 - x - 6}$

Factor the denominator: $x^2 - x - 6 = (x - 3)(x + 2)$ Now the limit becomes: $\lim_{x \to 3} \frac{2(x - 3)}{(x - 3)(x + 2)}$ Cancel the $x - 3$ terms: $\lim_{x \to 3} \frac{2}{x + 2} = \frac{2}{3 + 2} = \frac{2}{5}$

6. $\lim_{x \to -4} \frac{x^2 + 3x - 4}{x^2 + 7x + 12}$

Factor both the numerator and denominator: $x^2 + 3x - 4 = (x - 1)(x + 4) \quad \text{and} \quad x^2 + 7x + 12 = (x + 3)(x + 4)$ Now the limit becomes: $\lim_{x \to -4} \frac{(x - 1)(x + 4)}{(x + 3)(x + 4)}$ Cancel the $x + 4$ terms: $\lim_{x \to -4} \frac{x - 1}{x + 3} = \frac{-4 - 1}{-4 + 3} = \frac{-5}{-1} = 5$

Would you like further details on any step? Here are some related questions:

1. What is the general technique for factoring polynomials when evaluating limits?
2. How can you apply L'Hopital's Rule to limits like these when direct substitution gives indeterminate forms?
3. Why is it important to factor both the numerator and the denominator before canceling terms?
4. How do you handle limits involving higher-order polynomials or more complex expressions?
5. What are some common mistakes students make when solving limits involving factoring?

Tip: Always try direct substitution first to check for indeterminate forms before factoring.