Let `f(x)=[x] and g(x)={0, x in Z x^2, x in R -Z` then (where [.]denotest greatest integer funtion)

Options

A. `underset(x rarr 1)(lim) g(x)` exists, but g(x) is not continuous at x = 1
B. `underset(x rarr 1)(lim) f(x)` does not exist and f(x) is not continuous at x = 1
C. gof is continuous for all x.
D. fog is continuous for all x.

Answer

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

- \( f(x) = [x] \) (the greatest integer function)
- \( g(x) = \begin{cases} 0 & \text{if } x \in \mathbb{Z} \\ x^2 & \text{if } x \in \mathbb{R} - \mathbb{Z} \end{cases} \)

### Step 1: Analyze the continuity of \( f(x) = [x] \)

1. **Identify the points of discontinuity**: The greatest integer function \( f(x) = [x] \) is discontinuous at integer values of \( x \). To check this, we can evaluate the limits as \( x \) approaches an integer from the left and right.

2. **Evaluate at \( x = 1 \)**:
- \( \lim_{x \to 1^-} f(x) = [1 - \epsilon] = 0 \) (for any small \( \epsilon > 0 \))
- \( \lim_{x \to 1^+} f(x) = [1 + \epsilon] = 1 \)

3. **Conclusion for \( f(x) \)**: Since \( \lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x) \), \( f(x) \) is not continuous at \( x = 1 \).

### Step 2: Analyze the continuity of \( g(x) \)

1. **Evaluate at \( x = 1 \)**:
- For \( x = 1 \) (which is an integer), \( g(1) = 0 \).
- For \( x \) approaching 1 from the left (\( x \to 1^- \)):
- \( g(x) = x^2 \) since \( x \) is not an integer.
- \( \lim_{x \to 1^-} g(x) = 1^2 = 1 \).
- For \( x \) approaching 1 from the right (\( x \to 1^+ \)):
- Again, \( g(x) = x^2 \).
- \( \lim_{x \to 1^+} g(x) = 1^2 = 1 \).

2. **Conclusion for \( g(x) \)**:
- \( \lim_{x \to 1^-} g(x) = 1 \)
- \( \lim_{x \to 1^+} g(x) = 1 \)
- Since \( g(1) = 0 \), we have \( \lim_{x \to 1} g(x) \neq g(1) \). Thus, \( g(x) \) is not continuous at \( x = 1 \).

### Final Conclusion

- \( f(x) \) is not continuous at \( x = 1 \).
- \( g(x) \) is also not continuous at \( x = 1 \).

### Summary of Results
1. \( f(x) \) is discontinuous at \( x = 1 \).
2. \( g(x) \) is discontinuous at \( x = 1 \).

---