Your solution is thorough and clear, showcasing step-by-step factorization and substitution to evaluate the given limit. Below is a structured walkthrough and a summary of what we achieved:

Steps Recap:

1. Analyze the Numerator:

   • The numerator $x^4 - 9x^2 + 4x - 12$ was factored by grouping: $x^4 - 9x^2 + 4x - 12 = (x^2 - 9)(x^2 + 1) + 4(x - 3)$ Then $x^2 - 9$ was further factored as: $x^2 - 9 = (x - 3)(x + 3)$

2. Factorize the Denominator:

   • The denominator already contains a factor of $(x - 3)$ and $(4 - x)$.

3. Simplify the Expression:

   • The numerator was rewritten with $(x - 3)$ factored out: $(x - 3)[(x + 3)(x^2 + 1) + 4]$
   • This allowed canceling $(x - 3)$ (valid because $x \neq 3$).

4. Evaluate the Simplified Limit:

   • After canceling, the limit became: $\lim_{x \to 3} \frac{(x + 3)(x^2 + 1) + 4}{4 - x}$
   • Substituting $x = 3$, the numerator became: $(3 + 3)(3^2 + 1) + 4 = 6(9 + 1) + 4 = 60 + 4 = 64$ The denominator became: $4 - 3 = 1$

5. Final Answer:

   • The value of the limit is: $\boxed{64}$

Key Learning Point:

The critical step in solving this limit was recognizing the need to factorize the numerator to simplify and cancel the $(x - 3)$ term, avoiding an indeterminate $\frac{0}{0}$ form. Proper substitution followed after simplification.

Questions for Exploration:

1. What happens if $x^4 - 9x^2 + 4x - 12$ cannot be factored easily? Could polynomial division work instead?
2. How does factoring $(x - 3)$ prevent division by zero in this context?
3. Why is it valid to cancel $(x - 3)$ even though the limit approaches $x = 3$?
4. How would the solution differ if the denominator included terms that don’t vanish as $x \to 3$?
5. Can this problem be solved using L'Hôpital's Rule? Why or why not?

Tip:

Always check for common factors when evaluating limits involving polynomials. It simplifies the process and avoids the indeterminate form $\frac{0}{0}$.