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The expression you have provided involves evaluating the limit:

\[
\lim_{{x \to 1}} \left( 2 - \frac{x^2 + 2x - 3}{|x - 1|} \right)
\]

### Step-by-Step Solution:

1. **Factor the quadratic expression in the numerator**:
   The expression in the numerator is \( x^2 + 2x - 3 \). We factor it:
   \[
   x^2 + 2x - 3 = (x - 1)(x + 3)
   \]

2. **Rewrite the limit**:
   Substituting the factored form of the numerator:
   \[
   \lim_{{x \to 1}} \left( 2 - \frac{(x - 1)(x + 3)}{|x - 1|} \right)
   \]
   
3. **Simplify the absolute value expression**:
   Now, we consider the behavior of \( |x - 1| \) as \( x \to 1 \). The absolute value \( |x - 1| \) behaves differently depending on whether \( x \) approaches 1 from the left (\( x \to 1^- \)) or from the right (\( x \to 1^+ \)):

   - **As \( x \to 1^- \) (from the left)**, \( x - 1 < 0 \), so \( |x - 1| = -(x - 1) \).
   - **As \( x \to 1^+ \) (from the right)**, \( x - 1 > 0 \), so \( |x - 1| = x - 1 \).

4. **Evaluate left-hand and right-hand limits**:

   - **For \( x \to 1^+ \)**:
     When approaching from the right, we replace \( |x - 1| \) with \( (x - 1) \):
     \[
     \lim_{{x \to 1^+}} \left( 2 - \frac{(x - 1)(x + 3)}{x - 1} \right) = \lim_{{x \to 1^+}} \left( 2 - (x + 3) \right)
     \]
     Simplifying:
     \[
     = \lim_{{x \to 1^+}} (2 - x - 3) = 2 - 1 - 3 = -2
     \]

   - **For \( x \to 1^- \)**:
     When approaching from the left, we replace \( |x - 1| \) with \( -(x - 1) \):
     \[
     \lim_{{x \to 1^-}} \left( 2 - \frac{(x - 1)(x + 3)}{-(x - 1)} \right) = \lim_{{x \to 1^-}} \left( 2 + (x + 3) \right)
     \]
     Simplifying:
     \[
     = \lim_{{x \to 1^-}} (2 + x + 3) = 2 + 1 + 3 = 6
     \]

5. **Conclusion**:
   Since the left-hand limit (\( 6 \)) and the right-hand limit (\( -2 \)) are not equal, the limit does **not exist**.

### Final Answer:
The limit does not exist.

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Would you like further details or have any other questions?

Here are some related questions to explore:
1. What is the difference between left-hand and right-hand limits?
2. How does the absolute value function affect limit calculations?
3. How do we handle piecewise functions in limits?
4. What happens if the left-hand and right-hand limits are equal?
5. Can you give an example of a function where the limit exists but includes absolute values?

**Tip**: When working with limits involving absolute values, always consider the behavior of the function from both sides of the point where the absolute value changes.

Evaluate the limit as x approaches 1 of the expression: 2 - (x^2 + 2x - 3) / |x - 1|.

This problem focuses on evaluating the limit of a function involving absolute values and a quadratic expression as x approaches 1. It includes factoring the quadratic, analyzing left-hand and right-hand limits, and determining whether the overall limit exists. Suitable for advanced high school or college-level students.

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