`lim_(xtoc)f(x)` does not exist when
where `[.]` and `{.}` denotes greatest integer and fractional part of `x`

Options

A. `f(x)=[x]-[2x-1],c=3`
B. `f(x)=[x]-x,c=1`
C. `f(x)={x}^(2)-{-x}^(2),c=0`
D. `f(x)=(tan(sgnx))/((sgnx)),c=0`

Answer

To solve the problem of determining when the limit \( \lim_{x \to c} f(x) \) does not exist, we will analyze the given functions one by one. The functions involve the greatest integer function (denoted as \( [x] \)) and the fractional part function (denoted as \( \{x\} \)).

### Step-by-Step Solution

1. **Function Analysis for \( f(x) = [x] - [2x] - 1 \)**

- **Left-Hand Limit (LHL)**:
\[ \lim_{h \to 0} f(3 - h) = [3 - h] - [2(3 - h)] - 1 \]
- For \( h \) approaching \( 0 \) from the left, \( [3 - h] = 2 \) and \( [6 - 2h] = 5 \).
- Therefore, \( LHL = 2 - 5 - 1 = -4 \).

- **Right-Hand Limit (RHL)**:
\[ \lim_{h \to 0} f(3 + h) = [3 + h] - [2(3 + h)] - 1 \]
- For \( h \) approaching \( 0 \) from the right, \( [3 + h] = 3 \) and \( [6 + 2h] = 6 \).
- Therefore, \( RHL = 3 - 6 - 1 = -4 \).

- Since \( LHL = RHL \), the limit exists for this function.

2. **Function Analysis for \( f(x) = \{x\} - x \)** at \( c = 1 \)

- **Left-Hand Limit (LHL)**:
\[ \lim_{x \to 1^-} (\{x\} - x) = \{1\} - 1 = 0 - 1 = -1 \]

- **Right-Hand Limit (RHL)**:
\[ \lim_{x \to 1^+} (\{x\} - x) = \{1\} - 1 = 1 - 1 = 0 \]

- Since \( LHL \neq RHL \), the limit does not exist for this function.

3. **Function Analysis for \( f(x) = \{x\}^2 - \{-x\}^2 \)** at \( c = 0 \)

- **Left-Hand Limit (LHL)**:
\[ \lim_{x \to 0^-} (\{x\}^2 - \{-x\}^2) = (0)^2 - (1)^2 = 0 - 1 = -1 \]

- **Right-Hand Limit (RHL)**:
\[ \lim_{x \to 0^+} (\{x\}^2 - \{-x\}^2) = (0)^2 - (0)^2 = 0 - 0 = 0 \]

- Since \( LHL \neq RHL \), the limit does not exist for this function.

4. **Function Analysis for \( f(x) = \frac{\tan(\text{sgn}(x))}{\text{sgn}(x)} \)** at \( c = 0 \)

- **Left-Hand Limit (LHL)**:
\[ \lim_{x \to 0^-} \frac{\tan(-1)}{-1} = \tan(-1) \]

- **Right-Hand Limit (RHL)**:
\[ \lim_{x \to 0^+} \frac{\tan(1)}{1} = \tan(1) \]

- Since \( LHL = RHL \), the limit exists for this function.

### Conclusion

The limits do not exist for the functions in parts B and C. Therefore, the answer is:

- The limit \( \lim_{x \to c} f(x) \) does not exist when \( c = 1 \) (for \( f(x) = \{x\} - x \)) and \( c = 0 \) (for \( f(x) = \{x\}^2 - \{-x\}^2 \)).