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The problem you provided is asking to compute the following limit:

\[
\lim_{x \to 3} \frac{2x^3 - 6x^2 + 4x - 12}{x - 3}
\]

### Step-by-Step Solution:

1. **Substitute \(x = 3\) into the function:**
   
   Substituting \(x = 3\) directly into both the numerator and the denominator:
   
   \[
   \frac{2(3)^3 - 6(3)^2 + 4(3) - 12}{3 - 3} = \frac{54 - 54 + 12 - 12}{0} = \frac{0}{0}
   \]
   
   This gives us an indeterminate form \( \frac{0}{0} \), so we must apply **factoring** or **L'Hôpital's Rule**.

2. **Factor the numerator:**
   
   The expression in the numerator is a cubic polynomial. Let's factor it:

   First, factor \(x - 3\) from the cubic expression using synthetic division or factoring techniques:
   
   \[
   2x^3 - 6x^2 + 4x - 12 = (x - 3)(2x^2 + 0x + 4)
   \]
   
   So the limit simplifies to:
   
   \[
   \lim_{x \to 3} \frac{(x - 3)(2x^2 + 4)}{x - 3}
   \]

   We can now cancel the \(x - 3\) terms:
   
   \[
   \lim_{x \to 3} 2x^2 + 4
   \]

3. **Substitute \(x = 3\) into the simplified expression:**
   
   \[
   2(3)^2 + 4 = 2(9) + 4 = 18 + 4 = 22
   \]

### Final Answer:
The value of the limit is **22**.

Would you like any further details or have any questions? Here are five related questions you can consider:

1. How do you apply L'Hôpital's Rule to limits that result in \( \frac{0}{0} \)?
2. What are the methods for factoring polynomials to simplify limits?
3. Can every indeterminate form \( \frac{0}{0} \) be solved using L'Hôpital's Rule?
4. What happens if a cubic polynomial cannot be factored easily?
5. How do limits at infinity behave differently from limits at a finite value?

**Tip:** When faced with a \( \frac{0}{0} \) indeterminate form, try factoring or applying L'Hôpital’s Rule to simplify the expression before re-evaluating the limit.

lim (2x^3 - 6x^2 + 4x - 12) / (x - 3) as x -> 3

Learn how to compute the limit of (2x^3 - 6x^2 + 4x - 12) / (x - 3) as x approaches 3. This problem covers key calculus concepts such as factoring cubic polynomials and handling indeterminate forms using L'Hôpital's Rule. Suitable for high school students.

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