Solution:

(a) $\lim_{x \to 2} \frac{x^2 - x - 2}{x(x - 2)}$

1. Factorize the numerator: $x^2 - x - 2 = (x - 2)(x + 1)$

   So the expression becomes: $\frac{x^2 - x - 2}{x(x - 2)} = \frac{(x - 2)(x + 1)}{x(x - 2)}$

2. Simplify the expression: Cancel $(x - 2)$ (valid because $x \neq 2$): $\frac{(x - 2)(x + 1)}{x(x - 2)} = \frac{x + 1}{x}$

3. Evaluate the limit: Substitute $x = 2$: $\lim_{x \to 2} \frac{x + 1}{x} = \frac{2 + 1}{2} = \frac{3}{2}$

   Answer: $\frac{3}{2}$

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(b) $\lim_{x \to 3} \frac{x^3 - x^2 - 9x + 9}{x^2 - x - 6}$

1. Factorize the numerator: Group terms and factorize: $x^3 - x^2 - 9x + 9 = x^2(x - 1) - 9(x - 1) = (x - 1)(x^2 - 9)$

   Further factorize $x^2 - 9$ using the difference of squares: $x^3 - x^2 - 9x + 9 = (x - 1)(x - 3)(x + 3)$

2. Factorize the denominator: $x^2 - x - 6 = (x - 3)(x + 2)$

3. Simplify the expression: Substitute the factored forms: $\frac{x^3 - x^2 - 9x + 9}{x^2 - x - 6} = \frac{(x - 1)(x - 3)(x + 3)}{(x - 3)(x + 2)}$

   Cancel $(x - 3)$ (valid because $x \neq 3$): $\frac{(x - 1)(x + 3)}{x + 2}$

4. Evaluate the limit: Substitute $x = 3$: $\lim_{x \to 3} \frac{(x - 1)(x + 3)}{x + 2} = \frac{(3 - 1)(3 + 3)}{3 + 2} = \frac{2 \cdot 6}{5} = \frac{12}{5}$

   Answer: $\frac{12}{5}$

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Final Answers:

(a) $\frac{3}{2}$
(b) $\frac{12}{5}$

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Would you like detailed steps for a specific part or further clarifications?

Relative Questions:

1. How do we know when to factorize terms in a limit problem?
2. What happens if direct substitution leads to a valid result?
3. Why is it important to consider the domain when simplifying expressions?
4. What are common techniques to deal with indeterminate forms in limits?
5. Can L'Hopital's Rule be used here, and if so, how?

Tip: Always check for factorization opportunities when limits lead to indeterminate forms like $\frac{0}{0}$.