`lim_(xrarrc)f(x)` does not exist for
wher `[.]` represent greatest integer function `{.}` represent fractional part function

Options

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

Explanation

To solve the problem of determining when the limit \( \lim_{x \to c} f(x) \) does not exist for the given functions, we will analyze each option step by step.

### Step 1: Analyze Option A
Let \( f(x) = \lfloor x \rfloor - x \), where \( \lfloor x \rfloor \) is the greatest integer function.

**Find the limit as \( x \to 0 \):**
1. **Left-Hand Limit (LHL)**:
\[ \lim_{h \to 0^-} f(0 - h) = \lim_{h \to 0^-} \left( \lfloor -h \rfloor - (-h) \right) \]
For small positive \( h \), \( -h \) is slightly less than 0, so \( \lfloor -h \rfloor = -1 \).
\[ = -1 + h \to -1 \text{ as } h \to 0^- \]

2. **Right-Hand Limit (RHL)**:
\[ \lim_{h \to 0^+} f(0 + h) = \lim_{h \to 0^+} \left( \lfloor h \rfloor - h \right) \]
For small positive \( h \), \( \lfloor h \rfloor = 0 \).
\[ = 0 - h \to 0 \text{ as } h \to 0^+ \]

Since LHL \( \neq \) RHL, the limit does not exist at \( c = 0 \).

### Step 2: Analyze Option B
Let \( f(x) = \lfloor |x| \rfloor - \lfloor 2x - 1 \rfloor \).

**Find the limit as \( x \to 3 \):**
1. **Left-Hand Limit (LHL)**:
\[ \lim_{h \to 0^-} f(3 - h) = \lim_{h \to 0^-} \left( \lfloor 3 - h \rfloor - \lfloor 2(3 - h) - 1 \rfloor \right) \]
For \( h \) slightly less than 0, \( \lfloor 3 - h \rfloor = 3 \) and \( \lfloor 5 - 2h \rfloor = 4 \).
\[ = 3 - 4 = -1 \]

2. **Right-Hand Limit (RHL)**:
\[ \lim_{h \to 0^+} f(3 + h) = \lim_{h \to 0^+} \left( \lfloor 3 + h \rfloor - \lfloor 2(3 + h) - 1 \rfloor \right) \]
For \( h \) slightly greater than 0, \( \lfloor 3 + h \rfloor = 3 \) and \( \lfloor 5 + 2h \rfloor = 5 \).
\[ = 3 - 5 = -2 \]

Since LHL \( \neq \) RHL, the limit does not exist at \( c = 3 \).

### Step 3: Analyze Option C
Let \( f(x) = \{ x^2 \} - \{ -x^2 \} \), where \( \{ x \} \) is the fractional part function.

**Find the limit as \( x \to 0 \):**
1. **Left-Hand Limit (LHL)**:
\[ \lim_{h \to 0^-} f(0 - h) = \lim_{h \to 0^-} \left( \{ (-h)^2 \} - \{ -(-h)^2 \} \right) \]
Since \( (-h)^2 = h^2 \), we have:
\[ = \{ h^2 \} - \{ -h^2 \} = h^2 - (1 - h^2) = 2h^2 - 1 \]
As \( h \to 0^- \), this approaches \( -1 \).

2. **Right-Hand Limit (RHL)**:
\[ \lim_{h \to 0^+} f(0 + h) = \lim_{h \to 0^+} \left( \{ h^2 \} - \{ -h^2 \} \right) \]
Similarly, we have:
\[ = h^2 - (1 - h^2) = 2h^2 - 1 \]
As \( h \to 0^+ \), this also approaches \( -1 \).

Since LHL = RHL, the limit exists at \( c = 0 \).

### Step 4: Analyze Option D
Let \( f(x) = \frac{\tan(\text{sgn}(x))}{\text{sgn}(x)} \).

**Find the limit as \( x \to 0 \)**:
1. **Left-Hand Limit (LHL)**:
\[ \lim_{h \to 0^-} f(0 - h) = \lim_{h \to 0^-} \frac{\tan(-1)}{-1} = -\tan(1) \]

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

Since LHL \( \neq \) RHL, the limit does not exist at \( c = 0 \).

### Conclusion
The limits do not exist for options A and C.

### Final Answer
The limits do not exist for options A and C.