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The given mathematical expression is:

\[
\lim_{{x \to 2}} \frac{|x - 2|}{x - 2}
\]

This is a piecewise function, so to evaluate the limit, we need to consider the behavior of the expression as \(x\) approaches 2 from both the left and the right.

### Case 1: \(x \to 2^-\) (Approaching from the left)
When \(x < 2\), the absolute value function \(|x - 2|\) becomes \(-(x - 2)\). So the expression becomes:

\[
\frac{|x - 2|}{x - 2} = \frac{-(x - 2)}{x - 2} = -1
\]

Thus, as \(x\) approaches 2 from the left, the value of the function is \(-1\).

### Case 2: \(x \to 2^+\) (Approaching from the right)
When \(x > 2\), the absolute value function \(|x - 2|\) simplifies to \(x - 2\). So the expression becomes:

\[
\frac{|x - 2|}{x - 2} = \frac{(x - 2)}{x - 2} = 1
\]

Thus, as \(x\) approaches 2 from the right, the value of the function is \(1\).

### Conclusion:
Since the left-hand limit is \(-1\) and the right-hand limit is \(1\), the two limits do not agree. Therefore, the limit **does not exist**:

\[
\lim_{{x \to 2}} \frac{|x - 2|}{x - 2} \quad \text{does not exist}.
\]

Would you like further details or explanations? Here are 5 related questions to explore:

1. What is a one-sided limit and how does it differ from a two-sided limit?
2. How does the absolute value function affect limits?
3. Can limits exist at points of discontinuity?
4. How would this limit change if the expression were slightly modified, e.g., \( \lim_{{x \to 3}} \frac{|x - 3|}{x - 3} \)?
5. What is the relationship between continuity and the existence of limits?

**Tip:** When dealing with limits involving absolute values, always check the behavior from both sides of the point in question.

Evaluate the limit: \( \lim_{{x \to 2}} \frac{|x - 2|}{x - 2} \)

Discover how to evaluate the limit of the function \( \lim_{{x \to 2}} \frac{|x - 2|}{x - 2} \). This problem involves key concepts such as absolute value, piecewise functions, and left-hand and right-hand limits. Suitable for students in Grades 11-12.

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