if `f(x)=x^2sin(1/x) , x!=0` then the value of the function `f` at `x=0` so that the function is continuous at `x=0`

Options

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

Explanation

To find the value of the function \( f(x) \) at \( x = 0 \) such that the function is continuous at \( x = 0 \), we need to evaluate the limit of \( f(x) \) as \( x \) approaches 0.

Given:
\[ f(x) = x^2 \sin\left(\frac{1}{x}\right), \quad x \neq 0 \]

We want to find \( f(0) \) such that \( f \) is continuous at \( x = 0 \). For continuity at \( x = 0 \), we need:
\[ \lim_{x \to 0} f(x) = f(0) \]

### Step 1: Evaluate the limit as \( x \) approaches 0
We need to calculate:
\[ \lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) \]

### Step 2: Analyze the sine function
As \( x \) approaches 0, \( \frac{1}{x} \) approaches infinity. The sine function oscillates between -1 and 1:
\[ -1 \leq \sin\left(\frac{1}{x}\right) \leq 1 \]

### Step 3: Multiply by \( x^2 \)
Now, we can multiply the inequality 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 4: Apply the Squeeze Theorem
As \( x \) approaches 0, both \( -x^2 \) and \( x^2 \) approach 0:
\[ \lim_{x \to 0} -x^2 = 0 \quad \text{and} \quad \lim_{x \to 0} x^2 = 0 \]

By the Squeeze Theorem:
\[ \lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0 \]

### Step 5: Set \( f(0) \)
Thus, for the function to be continuous at \( x = 0 \):
\[ f(0) = \lim_{x \to 0} f(x) = 0 \]

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