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The given expression is the limit:

\[
\lim_{x \to 1} \frac{|x - 1|}{x - 1}
\]

To evaluate this limit, we need to consider the behavior of the absolute value function around \(x = 1\). Specifically, we'll look at the limit from both sides: the left-hand limit (\(x \to 1^{-}\)) and the right-hand limit (\(x \to 1^{+}\)).

### Case 1: \( x \to 1^{+} \) (Right-hand limit)
When \( x > 1 \), the absolute value \(|x - 1|\) simplifies to \(x - 1\). So the expression becomes:
\[
\frac{|x - 1|}{x - 1} = \frac{x - 1}{x - 1} = 1
\]
Thus, 
\[
\lim_{x \to 1^{+}} \frac{|x - 1|}{x - 1} = 1
\]

### Case 2: \( x \to 1^{-} \) (Left-hand limit)
When \( x < 1 \), the absolute value \(|x - 1|\) simplifies to \( -(x - 1) \). So the expression becomes:
\[
\frac{|x - 1|}{x - 1} = \frac{-(x - 1)}{x - 1} = -1
\]
Thus,
\[
\lim_{x \to 1^{-}} \frac{|x - 1|}{x - 1} = -1
\]

### Conclusion:
The left-hand and right-hand limits do not match:
- \( \lim_{x \to 1^{+}} = 1 \)
- \( \lim_{x \to 1^{-}} = -1 \)

Since these limits are different, the two-sided limit does **not exist**.

Would you like further details on this problem or have any questions? 

Here are some related questions you might explore:
1. What are the conditions for the existence of a two-sided limit?
2. How does the absolute value function affect the behavior of limits?
3. What is the difference between a one-sided limit and a two-sided limit?
4. Can you explain the geometric interpretation of absolute value in limits?
5. How do we handle limits where discontinuities are present?

**Tip:** Always break down absolute value limits into one-sided limits to check if they lead to different values!

Evaluate the limit: lim(x -> 1) |x - 1| / (x - 1)

This detailed solution explains how to evaluate the limit lim(x -> 1) |x - 1| / (x - 1) by analyzing both the left-hand and right-hand limits. The problem involves key concepts like absolute value and one-sided limits, suitable for students in Grades 11-12.

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