Let `f(x)=[x]+[-x]`, where `[x]` denotes the greastest integer less than or equal to x . Then, for any integer m

Options

A. `lim_(xtom) f(x)=f(m)`
B. `lim_(xtom)f(x)ne f(m)`
C. `lim_(xtom)f(x)` does not exist
D. none of these

Explanation

To solve the problem, we need to analyze the function \( f(x) = [x] + [-x] \), where \([x]\) denotes the greatest integer less than or equal to \( x \).

### Step-by-Step Solution:

1. **Understanding the Function**:
- The function \( f(x) \) is defined as \( f(x) = [x] + [-x] \).
- Here, \([x]\) is the greatest integer less than or equal to \( x \) and \([-x]\) is the greatest integer less than or equal to \(-x\).

2. **Case 1: When \( x \) is an Integer**:
- Let \( x = m \), where \( m \) is an integer.
- Then, \([m] = m\) and \([-m] = -m\).
- Therefore, \( f(m) = m + (-m) = 0 \).

3. **Case 2: When \( x \) is Not an Integer**:
- Let \( x \) be a non-integer, which can be expressed as \( x = m + \epsilon \) where \( 0 < \epsilon < 1 \).
- In this case, \([x] = m\) and \([-x] = -m - 1\) (since \(-x\) will be slightly less than \(-m\)).
- Thus, \( f(x) = [x] + [-x] = m + (-m - 1) = -1 \).

4. **Finding the Limits**:
- We need to evaluate the limits as \( x \) approaches \( m \) from both sides.
- **Limit as \( x \to m^+ \)** (approaching from the right):
- For \( x \to m^+ \), \( x \) is not an integer, hence \( f(x) = -1 \).
- Therefore, \( \lim_{x \to m^+} f(x) = -1 \).

- **Limit as \( x \to m^- \)** (approaching from the left):
- For \( x \to m^- \), \( x \) is also not an integer, hence \( f(x) = -1 \).
- Therefore, \( \lim_{x \to m^-} f(x) = -1 \).

5. **Comparing the Limits with \( f(m) \)**:
- We have \( f(m) = 0 \) (since \( m \) is an integer).
- The limits from both sides are \( -1 \).
- Since \( \lim_{x \to m^+} f(x) = -1 \) and \( \lim_{x \to m^-} f(x) = -1 \), but \( f(m) = 0 \), we conclude that:
\[ \lim_{x \to m} f(x) \neq f(m) \]

### Conclusion:
The function \( f(x) \) is not continuous at integer points \( m \). Therefore, the correct option is that the limit exists but does not equal the function value at that point.