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To solve the given problem, let's break down each part of the question.

We are given that \( f(x) = \lfloor x \rfloor + \lfloor -x \rfloor \), where \( \lfloor x \rfloor \) represents the greatest integer less than or equal to \( x \).

The goal is to show that the limit \( \lim_{x \to 2} f(x) \) exists, but it is not equal to \( f(2) \).

### Step 1: Understanding \( f(x) \)
The function is defined as:
\[
f(x) = \lfloor x \rfloor + \lfloor -x \rfloor
\]

For any integer \( a \), \( f(a) = \lfloor a \rfloor + \lfloor -a \rfloor \). Now, let's explore the function behavior near \( x = 2 \).

### Step 2: Evaluating the limit \( \lim_{x \to 2^-} f(x) \)
For \( x \to 2^- \) (values of \( x \) slightly less than 2), we have:
- \( \lfloor x \rfloor = 1 \) (since \( x < 2 \) but close to 2),
- \( \lfloor -x \rfloor = -2 \) (since \( -x \) is slightly greater than -2).

Thus, \( f(x) = 1 + (-2) = -1 \) when \( x \to 2^- \).

\[
\lim_{x \to 2^-} f(x) = -1
\]

### Step 3: Evaluating the limit \( \lim_{x \to 2^+} f(x) \)
For \( x \to 2^+ \) (values of \( x \) slightly greater than 2), we have:
- \( \lfloor x \rfloor = 2 \) (since \( x \) is now just above 2),
- \( \lfloor -x \rfloor = -3 \) (since \( -x \) is slightly less than -2).

Thus, \( f(x) = 2 + (-3) = -1 \) when \( x \to 2^+ \).

\[
\lim_{x \to 2^+} f(x) = -1
\]

### Step 4: Conclusion about the limit
Since both the left-hand limit and the right-hand limit agree, we can conclude:
\[
\lim_{x \to 2} f(x) = -1
\]

### Step 5: Calculating \( f(2) \)
At \( x = 2 \):
- \( \lfloor 2 \rfloor = 2 \),
- \( \lfloor -2 \rfloor = -2 \).

Thus, \( f(2) = 2 + (-2) = 0 \).

### Step 6: Final Conclusion
We have shown that:
\[
\lim_{x \to 2} f(x) = -1 \quad \text{and} \quad f(2) = 0
\]

Therefore, \( \lim_{x \to 2} f(x) \neq f(2) \).

---

Would you like a further detailed explanation or clarification on any of the steps? Here are five follow-up questions related to this:

1. What happens to the function \( f(x) \) near other integer points?
2. How does the behavior of floor functions impact discontinuities in functions like this?
3. Why are there holes at integer points in the graph of \( f(x) \)?
4. Can we generalize this limit behavior to other values of \( x \) where \( x \) is an integer?
5. How does the floor function handle negative values differently from positive values?

**Tip:** When evaluating limits involving piecewise or discontinuous functions, always check both the left-hand and right-hand limits to see if they match.

If f(x) = ⌊x⌋ + ⌊−x⌋, show that lim x→2 f(x) exists but is not equal to f(2).

Explore how to solve the limit lim x→2 f(x) where f(x) = ⌊x⌋ + ⌊−x⌋. Understand how the limit exists but differs from the actual value of f(2). This step-by-step solution focuses on limits, piecewise functions, and the properties of floor functions. Suitable for advanced high school students in Grades 11-12.

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