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The problem asks to evaluate the following limit:

\[
\lim_{x \to 2} \frac{x^2 - 8x + 12}{x^2 + 4x - 12}.
\]

### Step-by-step Solution:

1. **Substitute \( x = 2 \) directly:**

   Substitute \( x = 2 \) into the expression:

   \[
   \frac{(2)^2 - 8(2) + 12}{(2)^2 + 4(2) - 12} = \frac{4 - 16 + 12}{4 + 8 - 12} = \frac{0}{0}.
   \]

   The result is an indeterminate form (\(\frac{0}{0}\)), so further simplification is needed.

2. **Factor the numerator and denominator:**

   - Factor the numerator \( x^2 - 8x + 12 \):

     \[
     x^2 - 8x + 12 = (x - 2)(x - 6).
     \]

   - Factor the denominator \( x^2 + 4x - 12 \):

     \[
     x^2 + 4x - 12 = (x - 2)(x + 6).
     \]

3. **Simplify the expression:**

   After factoring, the limit becomes:

   \[
   \frac{(x - 2)(x - 6)}{(x - 2)(x + 6)}.
   \]

   Cancel the common factor \((x - 2)\):

   \[
   \frac{x - 6}{x + 6}.
   \]

4. **Substitute \( x = 2 \) again:**

   Substitute \( x = 2 \) into the simplified expression:

   \[
   \frac{2 - 6}{2 + 6} = \frac{-4}{8} = -\frac{1}{2}.
   \]

### Final Answer:

\[
\boxed{-\frac{1}{2}}
\]

Would you like more details or have any questions?

### Related Questions:
1. How do you handle indeterminate forms in limits?
2. What are some techniques for factoring quadratic expressions?
3. How do you find limits involving rational functions?
4. What are the common mistakes to avoid when simplifying limits?
5. How do you use L'Hôpital's Rule to solve indeterminate limits?

### Tip:
Always try to factor and simplify before substituting values in limits to avoid indeterminate forms!

Evaluate the limit lim(x → 2) [(x^2 - 8x + 12) / (x^2 + 4x - 12)].

Learn how to evaluate the limit lim(x → 2) [(x^2 - 8x + 12) / (x^2 + 4x - 12)] using factoring techniques to simplify indeterminate forms. This detailed explanation is perfect for high school or college students studying calculus.

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