We are asked to find the value of the following limit:

$\lim_{x \to 3} \frac{x^4 - 9x^2 + 4x - 12}{(x-3)(4-x)}.$

Step 1: Factor the numerator and the denominator

The denominator is already factored:

$(x-3)(4-x).$

We can rewrite $(4 - x)$ as $-(x - 4)$, which simplifies the denominator to:

$-(x - 3)(x - 4).$

Now, let’s focus on factoring the numerator:

$x^4 - 9x^2 + 4x - 12.$

Step 2: Try substituting $x = 3$ directly into the expression

Before factoring, let's first check whether direct substitution leads to an indeterminate form.

Substitute $x = 3$ into the numerator:

$3^4 - 9(3^2) + 4(3) - 12 = 81 - 9(9) + 12 - 12 = 81 - 81 + 12 - 12 = 0.$

Substitute $x = 3$ into the denominator:

$(3 - 3)(3 - 4) = 0 \cdot (-1) = 0.$

Since both the numerator and denominator are zero at $x = 3$, the limit is in the indeterminate form $\frac{0}{0}$. This suggests we should attempt factoring the numerator or use L'Hopital's Rule.

Step 3: Factor the numerator

Let’s attempt to factor the numerator $x^4 - 9x^2 + 4x - 12$.

We can start by grouping terms:

$x^4 - 9x^2 + 4x - 12 = (x^4 - 9x^2) + (4x - 12).$

Factor each group:

$x^4 - 9x^2 = x^2(x^2 - 9) = x^2(x - 3)(x + 3),$ $4x - 12 = 4(x - 3).$

Thus, the numerator becomes:

$x^2(x - 3)(x + 3) + 4(x - 3).$

Now factor out $(x - 3)$:

$= (x - 3) \left[ x^2(x + 3) + 4 \right].$

Simplify the expression inside the brackets:

$x^2(x + 3) + 4 = x^3 + 3x^2 + 4.$

Thus, the numerator factors as:

$(x - 3)(x^3 + 3x^2 + 4).$

Step 4: Simplify the expression

Now substitute the factored numerator and denominator into the original expression:

$\frac{(x - 3)(x^3 + 3x^2 + 4)}{-(x - 3)(x - 4)}.$

Cancel the common factor of $(x - 3)$ from both the numerator and the denominator (note that this is valid as $x \neq 3$):

$\frac{x^3 + 3x^2 + 4}{-(x - 4)}.$

Step 5: Evaluate the limit

Now, we can directly substitute $x = 3$ into the simplified expression:

$\frac{3^3 + 3(3^2) + 4}{-(3 - 4)} = \frac{27 + 27 + 4}{-(-1)} = \frac{58}{1} = 58.$

Final Answer:

The value of the limit is:

$\boxed{58}.$

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Would you like to go over any details or ask any follow-up questions? Here are some related questions you could explore:

1. How can we use L'Hopital's Rule to evaluate limits like this?
2. What happens if we attempt to factor other types of polynomials in similar limits?
3. How do we determine when a limit results in an indeterminate form such as $\frac{0}{0}$?
4. What are some other ways to simplify rational functions for limits?
5. How do the behavior of polynomials and rational functions change near points of discontinuity?

Tip: When encountering $\frac{0}{0}$ indeterminate forms, factoring or applying L'Hopital's Rule are common techniques to resolve the limit.