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To solve the problem, we need to determine the limit of the function \( f(x) \) as \( x \) approaches 2.

Given the piecewise function:
\[
f(x) = \begin{cases} 
\frac{(x-2)^2 \sin x}{|x-2|} & \text{for } x \neq 2 \\
\pi & \text{for } x = 2 
\end{cases}
\]
We are asked to find \( \lim_{x \to 2} f(x) \).

### Step 1: Evaluate the limit from the left and right as \( x \) approaches 2

Since \( f(x) \) is defined differently for \( x = 2 \), we first focus on the part of the function where \( x \neq 2 \):
\[
f(x) = \frac{(x-2)^2 \sin x}{|x-2|}
\]
For \( x \neq 2 \), let's evaluate the limit as \( x \) approaches 2:

1. **Case 1: \( x > 2 \) (Right-hand limit)**

   When \( x > 2 \), \( |x - 2| = x - 2 \). Therefore, the function simplifies to:
   \[
   f(x) = \frac{(x-2)^2 \sin x}{x-2} = (x-2)\sin x
   \]
   Now, taking the limit as \( x \) approaches 2:
   \[
   \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (x-2)\sin x
   \]
   Since \( x - 2 \) approaches 0 as \( x \) approaches 2, and \( \sin x \) is bounded, this limit is:
   \[
   \lim_{x \to 2^+} f(x) = 0
   \]

2. **Case 2: \( x < 2 \) (Left-hand limit)**

   When \( x < 2 \), \( |x - 2| = 2 - x \). Therefore, the function simplifies to:
   \[
   f(x) = \frac{(x-2)^2 \sin x}{-(x-2)} = -(x-2)\sin x
   \]
   Taking the limit as \( x \) approaches 2:
   \[
   \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} -(x-2)\sin x
   \]
   Again, since \( x - 2 \) approaches 0, the limit is:
   \[
   \lim_{x \to 2^-} f(x) = 0
   \]

### Step 2: Compare the left-hand and right-hand limits

Both the left-hand limit \( \lim_{x \to 2^-} f(x) \) and the right-hand limit \( \lim_{x \to 2^+} f(x) \) are equal to 0.

### Conclusion

Since the left-hand limit and right-hand limit are equal, the limit exists and is equal to 0.

Thus, \( \lim_{x \to 2} f(x) = 0 \), so the correct answer is **(A) 0**.

Would you like further details on any part of this problem? 

Here are five related questions that could expand your understanding:

1. What is the significance of left-hand and right-hand limits in determining the overall limit?
2. How would the limit change if the function had a different form for \( x = 2 \)?
3. What are some common scenarios where a limit does not exist?
4. How does the behavior of \( \sin x \) near \( x = 2 \) influence the result?
5. Can the value of a function at a point affect the limit at that point?

**Tip:** When evaluating limits, always check both the left-hand and right-hand limits, especially when dealing with piecewise functions.

If f is the function defined above, then \( \lim_{x \to 2} f(x) \) is

Learn how to determine the limit of a piecewise function as x approaches 2. This problem involves evaluating the left-hand and right-hand limits and applying key concepts in trigonometry and limit theory. Ideal for students in Grades 11-12.

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