Let `f (x) =[x] and g (x) =0` when x is an integer and `g (x) =x ^(2)` when x is not an integer ([] is ghe greatest integer function) then:

Options

A. `lim _(xto1) g(x)` exists, but g (x) is not continous at `x=1`
B. `lim _(xto1) f (x)` does not exist
C. gog is continous for all x
D. fog is continous for all x

Answer

To solve the given problem, we need to analyze the functions \( f(x) \) and \( g(x) \) as defined in the question.

### Step 1: Understand the Functions
1. **Function \( f(x) \)**:
- \( f(x) = [x] \) where \([x]\) is the greatest integer function (also known as the floor function). This function returns the largest integer less than or equal to \( x \).

2. **Function \( g(x) \)**:
- \( g(x) = 0 \) when \( x \) is an integer.
- \( g(x) = x^2 \) when \( x \) is not an integer.

### Step 2: Analyze \( g(x) \)
- For integer values of \( x \): \( g(x) = 0 \).
- For non-integer values of \( x \): \( g(x) = x^2 \).
- The graph of \( g(x) \) will have points where it is \( 0 \) at every integer and will follow the curve of \( y = x^2 \) between those integers.

### Step 3: Check Continuity of \( g(x) \)
- At integer points, \( g(x) \) is \( 0 \), but as we approach these points from the left or right (non-integer values), \( g(x) \) approaches the square of the integer, which is not \( 0 \). Thus, \( g(x) \) is not continuous at integer points.

### Step 4: Evaluate Limits
1. **Limit as \( x \to 1 \)**:
- Left-hand limit (LHL): \( \lim_{x \to 1^-} g(x) = 1^2 = 1 \)
- Right-hand limit (RHL): \( \lim_{x \to 1^+} g(x) = 1^2 = 1 \)
- \( g(1) = 0 \) (since \( 1 \) is an integer).
- Since LHL and RHL are equal but not equal to \( g(1) \), \( g(x) \) exists but is not continuous at \( x = 1 \).

2. **Limit as \( x \to 0 \)**:
- Left-hand limit (LHL): \( \lim_{x \to 0^-} f(x) = 0 \)
- Right-hand limit (RHL): \( \lim_{x \to 0^+} f(x) = 0 \)
- \( f(0) = 0 \) (since \( [0] = 0 \)).
- Thus, \( f(x) \) is continuous at \( x = 0 \).

### Step 5: Evaluate \( g(g(x)) \) and \( f(g(x)) \)
- **For \( g(g(x)) \)**:
- If \( x \) is an integer, \( g(x) = 0 \), thus \( g(g(x)) = g(0) = 0 \).
- If \( x \) is not an integer, \( g(x) = x^2 \), and since \( x^2 \) is non-integer for non-integer \( x \), \( g(g(x)) = g(x^2) = (x^2)^2 = x^4 \).
- Since \( g(g(x)) \) has discontinuities at integers, it is not continuous.

- **For \( f(g(x)) \)**:
- If \( x \) is an integer, \( g(x) = 0 \) and \( f(g(x)) = f(0) = [0] = 0 \).
- If \( x \) is not an integer, \( g(x) = x^2 \), and \( f(g(x)) = f(x^2) = [x^2] \).
- Since \( [x^2] \) has discontinuities at integers, \( f(g(x)) \) is also not continuous.

### Conclusion
- \( g(x) \) exists but is not continuous at integers.
- \( f(x) \) is continuous at \( x = 0 \) but not at \( x = 1 \).
- \( g(g(x)) \) and \( f(g(x)) \) are not continuous.

### Final Answer
The correct options are:
- \( g(x) \) exists but is not continuous at integers.
- \( f(x) \) does not exist at certain points.