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 evaluate the limit of \( f(x) \) as \( x \) approaches 0.

### Step 1: Understanding Continuity
A function is continuous at a point if:
\[ \lim_{x \to c} f(x) = f(c) \]
In this case, we want to ensure that:
\[ \lim_{x \to 0} f(x) = f(0) \]

### Step 2: Finding the Limit as \( x \) Approaches 0
We need to calculate:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) \]
Since \( \sin\left(\frac{1}{x}\right) \) oscillates between -1 and 1 for all \( x \neq 0 \), we can bound the function:
\[ -1 \leq \sin\left(\frac{1}{x}\right) \leq 1 \]
Thus, we can multiply through by \( x^2 \) (which is non-negative for all \( x \)):
\[ -x^2 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2 \]

### Step 3: Applying the Squeeze Theorem
Now, we take the limit of the bounding functions as \( x \) approaches 0:
\[ \lim_{x \to 0} -x^2 = 0 \]
\[ \lim_{x \to 0} x^2 = 0 \]
By the Squeeze Theorem:
\[ \lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0 \]

### Step 4: Setting the Value of the Function at \( x = 0 \)
Since the limit exists and equals 0, we can define \( f(0) \) as:
\[ f(0) = 0 \]

### Conclusion
Thus, the value of the function at \( x = 0 \) is:
\[ \boxed{0} \]