The function f defined as - `f(x) = (sin x^(2))//x` for `x ne 0` and f(0) = 0 is:

Options

A. continuous, and derivable at x = 0
B. neither continuous nor derivable at x = 0
C. continuous but not derivable at x =.0
D. none of these

Answer

To analyze the function \( f(x) = \frac{\sin(x^2)}{x} \) for \( x \neq 0 \) and \( f(0) = 0 \), we need to determine the continuity of the function at \( x = 0 \). We will do this by checking if the limit of \( f(x) \) as \( x \) approaches 0 is equal to \( f(0) \).

### Step-by-Step Solution:

1. **Identify the function**:
\[ f(x) = \frac{\sin(x^2)}{x} \quad \text{for } x \neq 0 \]
\[ f(0) = 0 \]

2. **Find the limit as \( x \) approaches 0**:
We need to compute:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sin(x^2)}{x} \]

3. **Use substitution**:
Let \( y = x^2 \). Then as \( x \to 0 \), \( y \to 0 \) as well, and \( x = \sqrt{y} \). Thus, we can rewrite the limit:
\[ \lim_{x \to 0} \frac{\sin(x^2)}{x} = \lim_{y \to 0} \frac{\sin(y)}{\sqrt{y}} \]

4. **Apply L'Hôpital's Rule**:
Since both the numerator and denominator approach 0 as \( y \to 0 \), we can apply L'Hôpital's Rule:
\[ \lim_{y \to 0} \frac{\sin(y)}{\sqrt{y}} = \lim_{y \to 0} \frac{\cos(y)}{\frac{1}{2\sqrt{y}}} \]
This simplifies to:
\[ \lim_{y \to 0} 2\sqrt{y} \cos(y) \]

5. **Evaluate the limit**:
As \( y \to 0 \), \( \cos(y) \to 1 \) and \( \sqrt{y} \to 0 \):
\[ \lim_{y \to 0} 2\sqrt{y} \cos(y) = 2 \cdot 0 \cdot 1 = 0 \]

6. **Conclusion**:
Since:
\[ \lim_{x \to 0} f(x) = 0 = f(0) \]
Therefore, the function \( f(x) \) is continuous at \( x = 0 \).

### Final Answer:
The function \( f(x) \) is continuous at \( x = 0 \).