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

Answer

To solve the problem, we will analyze the function \( f(x) = \frac{[x] - [\sin x]}{1 - \sin [\cos x]} \) and the function \( g(x) = \frac{1}{f(x)} \). We need to determine the limits as \( x \) approaches 0 for both \( f(x) \) and \( g(x) \).

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

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

#### Sub-step 1.1: Calculate \( [x] \) as \( x \to 0 \)

As \( x \) approaches 0 from the left (\( x \to 0^- \)):
- \( [x] = 0 \) because the greatest integer less than or equal to 0 is 0.

As \( x \) approaches 0 from the right (\( x \to 0^+ \)):
- \( [x] = 0 \) for the same reason.

#### Sub-step 1.2: Calculate \( [\sin x] \)

As \( x \to 0 \):
- \( \sin x \to 0 \)
- Therefore, \( [\sin x] = 0 \).

#### Sub-step 1.3: Calculate \( [\cos x] \)

As \( x \to 0 \):
- \( \cos x \to 1 \)
- Therefore, \( [\cos x] = 1 \).

#### Sub-step 1.4: Calculate \( \sin [\cos x] \)

Since \( [\cos x] = 1 \):
- \( \sin [\cos x] = \sin(1) \).

#### Sub-step 1.5: Substitute values into \( f(x) \)

Now substituting these values into \( f(x) \):
\[ f(x) = \frac{0 - 0}{1 - \sin(1)} = \frac{0}{1 - \sin(1)} = 0 \]

### Step 2: Analyze the limits

#### Sub-step 2.1: Left-hand limit (LHL)

Calculate the left-hand limit as \( x \to 0^- \):
\[ \lim_{x \to 0^-} f(x) = 0 \]

#### Sub-step 2.2: Right-hand limit (RHL)

Calculate the right-hand limit as \( x \to 0^+ \):
\[ \lim_{x \to 0^+} f(x) = 0 \]

Since both limits exist and are equal:
\[ \lim_{x \to 0} f(x) = 0 \]

### Step 3: Analyze \( g(x) \)

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

Since \( f(x) \) approaches 0 as \( x \to 0 \), we need to evaluate the limit of \( g(x) \):
\[ \lim_{x \to 0} g(x) = \lim_{x \to 0} \frac{1}{f(x)} \]

Since \( f(x) \) approaches 0, \( g(x) \) approaches infinity:
\[ \lim_{x \to 0} g(x) = \infty \]

### Conclusion

1. \( \lim_{x \to 0} f(x) \) exists and equals 0.
2. \( \lim_{x \to 0} g(x) \) does not exist (it approaches infinity).