If f(x) = x² sin1x, where x ≠ 0, then the value of the function f(x) at x = 0, so that the function is continuous at x = 0 is

The correct option is 0.

To find the value of the function $f\left(x\right)$ at $x=0$ so that the function is continuous at $x=0$, we must apply the definition of continuity.

A function $f\left(x\right)$ is continuous at $x=0$ if and only if the limit of the function as $x$ approaches $0$ exists and is equal to the value of the function at $x=0$.
Mathematically, this condition is written as:

$$\underset{x\to 0}{lim}f\left(x\right)=f\left(0\right)$$

We are given that for $x\ne 0$, the function is defined as:

$$f\left(x\right)=x^{2}sin\left(\frac{1}{x}\right)$$

Let us evaluate the limit of $f\left(x\right)$ as $x$ approaches $0$:

$$L=\underset{x\to 0}{lim}{x}^{2}sin\left(\frac{1}{x}\right)$$

To evaluate this limit, we can use the Sandwich Theorem (or Squeeze Theorem). We know that the sine function is bounded between $-1$ and $1$ for all real numbers. Thus, for any $x\ne 0$:

$$-1\le sin\left(\frac{1}{x}\right)\le 1$$

Since $x^2>0$ for all $x\ne 0$, multiplying the entire inequality by $x^2$ preserves the inequality signs:

$$-x^2\le x^{2}sin\left(\frac{1}{x}\right)\le x^2$$

Now, we take the limit of the outer functions as $x$ approaches $0$:
Left limit:

$$\underset{x\to 0}{lim}(-x^2)=-{\left(0\right)}^2=0$$

Right limit:

$$\underset{x\to 0}{lim}\left(x^2\right)={\left(0\right)}^2=0$$

By the Sandwich Theorem, since the limits of both bounding functions as $x\to 0$ are equal to $0$, the limit of the middle function must also be $0$:

$$\underset{x\to 0}{lim}{x}^{2}sin\left(\frac{1}{x}\right)=0$$

Therefore, for the function $f\left(x\right)$ to be continuous at $x=0$, we must define:

$$f\left(0\right)=\underset{x\to 0}{lim}f\left(x\right)=0$$