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The problem asks to evaluate the following limits for the function \( f(x) = |x - 2| \).

1. **First Limit**: 
   \[
   \lim_{x \to 2^-} \frac{f(x) - f(2)}{x - 2}
   \]
   Here, \( f(x) = |x - 2| \) and \( f(2) = |2 - 2| = 0 \), so the expression becomes:
   \[
   \frac{|x - 2| - 0}{x - 2} = \frac{|x - 2|}{x - 2}
   \]
   When \( x \to 2^- \) (approaching from the left), \( x - 2 \) is negative, so \( |x - 2| = -(x - 2) \). Thus, the limit becomes:
   \[
   \lim_{x \to 2^-} \frac{-(x - 2)}{x - 2} = -1
   \]

2. **Second Limit**:
   \[
   \lim_{x \to 2^+} \frac{f(x) - f(2)}{x - 2}
   \]
   Again, using \( f(2) = 0 \), the expression simplifies to:
   \[
   \frac{|x - 2|}{x - 2}
   \]
   When \( x \to 2^+ \) (approaching from the right), \( x - 2 \) is positive, so \( |x - 2| = x - 2 \). The limit becomes:
   \[
   \lim_{x \to 2^+} \frac{x - 2}{x - 2} = 1
   \]

### Final Answers:
1. \( \lim_{x \to 2^-} \frac{f(x) - f(2)}{x - 2} = -1 \)
2. \( \lim_{x \to 2^+} \frac{f(x) - f(2)}{x - 2} = 1 \)

Since the two one-sided limits do not agree, the function \( f(x) = |x - 2| \) is **not differentiable** at \( x = 2 \).

Would you like more details or further clarification on any part of this solution?

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### Related Questions:
1. How do you formally define differentiability at a point?
2. What are the conditions for a piecewise function to be differentiable at a point?
3. Can a function be continuous but not differentiable? Provide examples.
4. What is the geometric interpretation of differentiability at a point?
5. How can we compute limits involving absolute value functions for different points?

**Tip**: When evaluating limits for absolute value functions, always consider the behavior of the function from both the left and right to handle cases where the absolute value expression changes form.

Let f(x) = |x - 2|. Evaluate the following limits:
1. lim (x -> 2-) [f(x) - f(2)] / (x - 2)
2. lim (x -> 2+) [f(x) - f(2)] / (x - 2)

Learn how to evaluate one-sided limits of the function f(x) = |x - 2| as x approaches 2. This problem explores the behavior of absolute value functions and their differentiability at a point. Step-by-step solutions for limits approaching from the left and right are provided. Suitable for high school students in Grades 11-12.

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