Let the function `f(x)=x^(2)sin((1)/(x)), AA x ne 0` is continuous at x = 0. Then, the vaue of the function at x = 0 is

Options

A. 0
B. `-1`
C. 1
D. indeterminate

Answer

To find the value of the function \( f(x) = x^2 \sin\left(\frac{1}{x}\right) \) at \( x = 0 \) such that the function is continuous at that point, we need to determine the limit of \( f(x) \) as \( x \) approaches 0.

### Step-by-Step Solution:

1. **Identify the function and the point of interest**:
The function is given as \( f(x) = x^2 \sin\left(\frac{1}{x}\right) \) for \( x \neq 0 \). We need to find \( f(0) \) such that \( f(x) \) is continuous at \( x = 0 \).

2. **Set up the limit**:
To ensure continuity at \( x = 0 \), we need to evaluate:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) \]

3. **Analyze the sine function**:
We know that the sine function is bounded:
\[ -1 \leq \sin\left(\frac{1}{x}\right) \leq 1 \]
Therefore, we can multiply through by \( x^2 \):
\[ -x^2 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2 \]

4. **Apply the Squeeze Theorem**:
As \( x \) approaches 0, both \( -x^2 \) and \( x^2 \) approach 0. Thus, by the Squeeze Theorem:
\[ \lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0 \]

5. **Conclude the value at \( x = 0 \)**:
Since the limit exists and equals 0, we can define:
\[ f(0) = 0 \]
This ensures that \( f(x) \) is continuous at \( x = 0 \).

### Final Answer:
The value of the function at \( x = 0 \) is \( \boxed{0} \).