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

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 find \( f(0) \) such that:

\[ \lim_{x \to 0} f(x) = f(0) \]

### Step 1: Evaluate the limit as \( x \) approaches 0

We start by calculating the limit:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) \]

### Step 2: Use the Squeeze Theorem

We know that the sine function is bounded:

\[ -1 \leq \sin\left(\frac{1}{x}\right) \leq 1 \]

Multiplying through by \( x^2 \) (which is non-negative for \( x \neq 0 \)) gives:

\[ -x^2 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2 \]

### Step 3: Apply the Squeeze Theorem

Now, we can apply the Squeeze Theorem. As \( x \) approaches 0, both \( -x^2 \) and \( x^2 \) approach 0:

\[ \lim_{x \to 0} -x^2 = 0 \quad \text{and} \quad \lim_{x \to 0} x^2 = 0 \]

Thus, by the Squeeze Theorem:

\[ \lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0 \]

### Step 4: Set \( f(0) \)

To ensure continuity at \( x = 0 \), we need:

\[ f(0) = \lim_{x \to 0} f(x) = 0 \]

### Conclusion

Therefore, the value of the function \( f \) at \( x = 0 \) should be:

\[ f(0) = 0 \]