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The limit expression provided is:

\[
\lim_{{x \to 2^+}} \frac{1 - x}{x - 2}
\]

We need to evaluate this limit as \( x \) approaches 2 from the right (\( x \to 2^+ \)).

### Step-by-Step Solution:
1. **Rewriting the Expression**:
   The denominator \( x - 2 \) approaches 0 as \( x \to 2 \). Specifically, since we are approaching from the right (\( x > 2 \)), \( x - 2 \) is slightly positive.

   The numerator \( 1 - x \) approaches \( 1 - 2 = -1 \) as \( x \to 2 \).

2. **Behavior Near \( x = 2 \)**:
   - Numerator: \( 1 - x \to -1 \)
   - Denominator: \( x - 2 \to 0^+ \) (slightly positive)

3. **Evaluating the Limit**:
   The expression simplifies to:

   \[
   \lim_{{x \to 2^+}} \frac{-1}{x - 2}
   \]

   As \( x \) approaches 2 from the right, \( x - 2 \) approaches a very small positive number, making the fraction \( \frac{-1}{x - 2} \) approach negative infinity:

   \[
   \lim_{{x \to 2^+}} \frac{-1}{x - 2} = -\infty
   \]

### Conclusion:
The limit is:

\[
\lim_{{x \to 2^+}} \frac{1 - x}{x - 2} = -\infty
\]

Would you like more details on any step? Here are some related questions:

1. What happens if we take the limit as \( x \to 2^- \) (from the left)?
2. How do you evaluate limits involving infinity?
3. Can this limit be solved using L'Hopital's Rule?
4. What is the significance of approaching a point from the left or the right in limits?
5. How can we apply this concept to piecewise functions?

**Tip**: Always consider the direction of approach when evaluating limits, especially when a term goes to zero.

Learn how to evaluate the right-hand limit lim (x -> 2+) (1 - x) / (x - 2), where the denominator approaches zero and the numerator leads to negative infinity. This problem introduces key concepts in calculus involving limits and singularities. Suitable for high school students in Grades 11-12.

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