If `f(x) = x^(2)sin'(1)/(x)`, where `x ne 0`, then the value of the function f at `x = 0`, so that the function is continuous at `x = 0` is

Options

A. 0
B. -1
C. 1
D. None of the above

Explanation

To determine 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 follow these steps:

### Step 1: Understand Continuity
A function is continuous at a point \( x = a \) if:
1. \( f(a) \) is defined.
2. \( \lim_{x \to a} f(x) \) exists.
3. \( \lim_{x \to a} f(x) = f(a) \).

In our case, we want to find \( f(0) \) such that \( f(x) \) is continuous at \( x = 0 \).

### Step 2: Find the Limit as \( x \) Approaches 0
We need to calculate \( \lim_{x \to 0} f(x) \):
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) \]

### Step 3: Analyze the Limit
As \( x \) approaches 0, \( \sin\left(\frac{1}{x}\right) \) oscillates between -1 and 1. Thus, we can bound \( f(x) \):
\[ -x^2 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2 \]

### Step 4: Apply the Squeeze Theorem
Since both \( -x^2 \) and \( x^2 \) approach 0 as \( x \) approaches 0, we can apply the Squeeze Theorem:
\[ \lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0 \]

### Step 5: Define \( f(0) \)
To ensure continuity at \( x = 0 \), we set:
\[ f(0) = \lim_{x \to 0} f(x) = 0 \]

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

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