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To solve the problem of finding \(\lim_{{x \to 2}} f(x)\) where the function \(f(x)\) is defined as:

\[
f(x) = \begin{cases} 
\frac{(x - 2)^2 \sin x}{|x - 2|} & \text{for } x \neq 2 \\
\pi & \text{for } x = 2 
\end{cases}
\]

### Step 1: Analyze the limit for \(x \neq 2\)
For \(x \neq 2\), the function is:
\[
f(x) = \frac{(x - 2)^2 \sin x}{|x - 2|}
\]

This can be simplified to:
\[
f(x) = |x - 2| \cdot (x - 2) \cdot \sin x
\]

### Step 2: Take the limit as \(x\) approaches 2
We are interested in:
\[
\lim_{{x \to 2}} f(x) = \lim_{{x \to 2}} |x - 2| \cdot (x - 2) \cdot \sin x
\]

### Step 3: Evaluate the components of the limit
1. \( |x - 2| \) approaches 0 as \( x \) approaches 2.
2. \( (x - 2) \) also approaches 0 as \( x \) approaches 2.
3. \( \sin x \) approaches \(\sin 2\), which is a finite number.

### Step 4: Conclusion
Since both \( |x - 2| \) and \( (x - 2) \) approach 0, the entire product approaches 0.

Therefore:
\[
\lim_{{x \to 2}} f(x) = 0
\]

### Step 5: Consider the given function at \( x = 2 \)
The function is defined as \( f(2) = \pi \). However, the limit depends on the function's behavior as \(x\) approaches 2 from both sides, not just the value at \(x = 2\).

Since the limit is determined by the function's behavior for \(x \neq 2\), we conclude:
\[
\lim_{{x \to 2}} f(x) = 0
\]

### Answer
The correct answer is \( \boxed{0} \) (Option A).

Would you like further details or explanations? 

Here are five questions related to this topic:

1. What is the difference between a limit and the actual value of a function at a point?
2. How does the concept of continuity relate to limits?
3. What would happen if the limit did not exist at \( x = 2 \)?
4. How do we determine limits for piecewise functions in general?
5. Can a limit exist even if the function is undefined at that point?

**Tip:** When approaching limit problems, always consider simplifying the expression before directly substituting the value to avoid indeterminate forms.

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

Solve a limit problem for a piecewise function with trigonometric components. Learn how to evaluate \(\lim_{{x \to 2}} f(x)\) for the function \(f(x)\), defined using sine and absolute values. Suitable for Grades 11-12 students.

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