Let `f(x) = ([x] - ["sin" x])/(1 - "sin" [cos x])` and `g (x) = (1)/(f(x))` and [] represents the greatest integer function, then

Options

A. (A)`underset(x to 0)(lim) f(x)` exist
B. (B)`underset(x to 0) (1)/(g(x))` exists
C. (C)`underset(x to 0)(lim) g (x)` does not exists
D. (D)`underset(x to 0)(lim) f(x) = 0`

Answer

To solve the problem, we need to analyze the functions \( f(x) \) and \( g(x) \) given by:

\[ f(x) = \frac{[x] - [\sin x]}{1 - \sin [\cos x]} \]

\[ g(x) = \frac{1}{f(x)} \]

where \( [x] \) denotes the greatest integer function (or floor function).

### Step 1: Analyze \( f(x) \)

1. **Evaluate \( f(x) \) as \( x \) approaches 0 from the left:**

For \( x \to 0^- \):
- \( [x] = 0 \) (since \( x \) is negative and less than 1)
- \( [\sin x] = 0 \) (since \( \sin x \) approaches 0)
- \( [\cos x] = 0 \) (since \( \cos x \) approaches 1)

Thus, we have:
\[ f(0^-) = \frac{0 - 0}{1 - \sin(0)} = \frac{0}{1 - 0} = 0 \]

### Step 2: Evaluate \( f(x) \) as \( x \) approaches 0 from the right:

2. **Evaluate \( f(x) \) as \( x \) approaches 0 from the right:**

For \( x \to 0^+ \):
- \( [x] = 0 \)
- \( [\sin x] = 0 \)
- \( [\cos x] = 0 \)

Thus, we have:
\[ f(0^+) = \frac{0 - 0}{1 - \sin(0)} = \frac{0}{1 - 0} = 0 \]

### Step 3: Confirm the limits

3. **Confirm the limits:**

Since both left-hand limit and right-hand limit of \( f(x) \) as \( x \to 0 \) are equal:
\[ \lim_{x \to 0} f(x) = 0 \]

### Step 4: Evaluate \( g(x) \)

4. **Evaluate \( g(x) \):**

Since \( g(x) = \frac{1}{f(x)} \), we find:
\[ g(0) = \frac{1}{f(0)} = \frac{1}{0} \]
This indicates that \( g(x) \) does not exist at \( x = 0 \).

### Step 5: Conclusion

5. **Summarize the findings:**
- \( \lim_{x \to 0} f(x) \) exists and equals 0.
- \( g(x) \) does not exist at \( x = 0 \).

### Final Answer

Based on the analysis, we conclude:
- The limit \( \lim_{x \to 0} f(x) \) exists and is equal to 0.
- The limit \( \lim_{x \to 0} g(x) \) does not exist.