If 
f(x) = [[x]] + [[−x]],
 show that 
lim 
x→2 
f(x)
 exists but is not equal to f(2).
The graph of 
f(x) = [[x]] + [[−x]]
 is the same as the graph of 
g(x) = −1
 with holes at each integer, since 
f(a) = 
 
 for any integer a. Also,
lim 
x→2− 
f(x) = 
 
  and  
lim 
x→2+ 
f(x) = 
  ,
  so  
lim 
x→2 
f(x) = 
  .
However,
f(2) = [[2]] + [[−2]] = 2 + 
  = 
  ,
so 
lim 
x→2 
f(x) ≠ f(2).

Let's work through this step by step to show the required properties about the function \( f(x) = \lfloor x \rfloor + \lfloor -x \rfloor \), where \( \lfloor \cdot \rfloor \) represents the greatest integer function or the floor function.

### Step 1: Analyze the behavior of \( f(x) \)
We know that:
\[
f(x) = \lfloor x \rfloor + \lfloor -x \rfloor
\]
The function \( \lfloor x \rfloor \) returns the greatest integer less than or equal to \( x \), and similarly, \( \lfloor -x \rfloor \) returns the greatest integer less than or equal to \( -x \).

For non-integer values of \( x \), this function exhibits a predictable pattern because:
- \( \lfloor x \rfloor \) jumps to the next integer as \( x \) crosses an integer,
- \( \lfloor -x \rfloor \) jumps similarly, but the pattern is inverted.

### Step 2: Behavior around \( x = 2 \)
To evaluate the limit, we first check the left-hand and right-hand limits at \( x = 2 \).

#### Left-hand limit (\( x \to 2^- \)):
Consider values of \( x \) slightly less than 2, e.g., \( x = 1.9 \):
\[
f(1.9) = \lfloor 1.9 \rfloor + \lfloor -1.9 \rfloor = \lfloor 1.9 \rfloor + \lfloor -1.9 \rfloor = 1 + (-2) = -1
\]
This will hold for any \( x \in (1, 2) \). Hence:
\[
\lim_{x \to 2^-} f(x) = -1
\]

#### Right-hand limit (\( x \to 2^+ \)):
Consider values of \( x \) slightly greater than 2, e.g., \( x = 2.1 \):
\[
f(2.1) = \lfloor 2.1 \rfloor + \lfloor -2.1 \rfloor = \lfloor 2.1 \rfloor + \lfloor -2.1 \rfloor = 2 + (-3) = -1
\]
This will hold for any \( x \in (2, 3) \). Hence:
\[
\lim_{x \to 2^+} f(x) = -1
\]

### Step 3: Conclude the limit exists
Since:
\[
\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = -1
\]
we conclude:
\[
\lim_{x \to 2} f(x) = -1
\]

### Step 4: Check the value at \( x = 2 \)
At \( x = 2 \), we directly compute \( f(2) \):
\[
f(2) = \lfloor 2 \rfloor + \lfloor -2 \rfloor = 2 + (-2) = 0
\]

### Step 5: Compare the limit with the value of \( f(2) \)
We found that:
\[
\lim_{x \to 2} f(x) = -1 \quad \text{and} \quad f(2) = 0
\]
Thus:
\[
\lim_{x \to 2} f(x) \neq f(2)
\]

### Conclusion
The limit \( \lim_{x \to 2} f(x) \) exists, but it is not equal to \( f(2) \), confirming that the limit exists but does not match the function's value at \( x = 2 \).

---

Would you like further details or have any questions?

Here are 5 related questions:
1. What are the properties of the floor function \( \lfloor x \rfloor \)?
2. Can you generalize this result for other integers \( a \) where \( x \to a \)?
3. How does the behavior of \( f(x) \) change as \( x \) approaches non-integer values?
4. What is the significance of a function having a limit that differs from its value at a point?
5. What other types of discontinuities exist besides jump discontinuities?

**Tip:** The floor function creates discontinuities at integer points, leading to potential mismatches between limits and function values at those points.

If f(x) = [[x]] + [[−x]], show that lim x→2 f(x) exists but is not equal to f(2). The graph of f(x) = [[x]] + [[−x]] is the same as the graph of g(x) = −1 with holes at each integer, since f(a) = for any integer a. Also, lim x→2− f(x) = and lim x→2+ f(x) = , so lim x→2 f(x) = . However, f(2) = [[2]] + [[−2]] = 2 + = , so lim x→2 f(x) ≠ f(2).

Explore the limit of the function f(x) = ⌊x⌋ + ⌊-x⌋ as x approaches 2. This problem demonstrates the behavior of the floor function and how a jump discontinuity leads to the limit existing but differing from the function's value at x = 2. Suitable for undergraduate mathematics students.

Math Problem Statement

and
lim x→2+ f(x) = , so
lim x→2 f(x) = . However, f(2) = [[2]] + [[−2]] = 2 +

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