If `f(x)={(x-1",",xlt0),(1/4",",x=0),(x^2",",xgt0):}` , then

Options

A. `underset(xrarr0+)(lim) f(x)=1`
B. `underset(xrarr0-)(lim) f(x)=1`
C. f (X) is discontinuous at x = 0
D. None of these

Answer

To determine whether the function \( f(x) \) is continuous at \( x = 0 \), we need to check the following conditions:

1. **Calculate \( f(0) \)**:
Given that \( f(0) = \frac{1}{4} \).

2. **Calculate the left-hand limit \( \lim_{x \to 0^-} f(x) \)**:
For \( x < 0 \), \( f(x) = x - 1 \).
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (x - 1) = 0 - 1 = -1 \]

3. **Calculate the right-hand limit \( \lim_{x \to 0^+} f(x) \)**:
For \( x > 0 \), \( f(x) = x^2 \).
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x^2) = 0^2 = 0 \]

4. **Check the continuity condition**:
A function \( f(x) \) is continuous at \( x = 0 \) if:
\[ \lim_{x \to 0^-} f(x) = f(0) = \lim_{x \to 0^+} f(x) \]
From our calculations:
- \( \lim_{x \to 0^-} f(x) = -1 \)
- \( f(0) = \frac{1}{4} \)
- \( \lim_{x \to 0^+} f(x) = 0 \)

Since \( -1 \neq \frac{1}{4} \) and \( \frac{1}{4} \neq 0 \), the function is not continuous at \( x = 0 \).

### Conclusion:
The function \( f(x) \) is not continuous at \( x = 0 \).

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