If `f(x)={{:(x^(2)"sin"(1)/(x)",",x ne0),(k",",x=0):}`

Options

A. 1
B. `-1`
C. 0
D. none of these

Answer

To find the value of \( k \) for the function defined as:

\[ f(x) = \begin{cases} x^2 \sin\left(\frac{1}{x}\right) & \text{if } x \neq 0 \\ k & \text{if } x = 0 \end{cases} \]

we need to ensure that the function is continuous at \( x = 0 \). For continuity at \( x = 0 \), we need to check the following condition:

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

### Step 1: Calculate \( f(0) \)

Since \( f(0) = k \), we have:

\[ f(0) = k \]

### Step 2: Calculate \( \lim_{x \to 0} f(x) \)

For \( x \neq 0 \), we have:

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

Now, we need to find:

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

### Step 3: Evaluate the limit

We know that \( \sin\left(\frac{1}{x}\right) \) oscillates between -1 and 1. Therefore, we can use the Squeeze Theorem to evaluate the limit:

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

As \( x \to 0 \), both \( -x^2 \) and \( x^2 \) approach 0. Thus, by the Squeeze Theorem:

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

### Step 4: Set the limit equal to \( f(0) \)

Now we have:

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

For continuity at \( x = 0 \):

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

### Conclusion

Thus, the value of \( k \) is:

\[ \boxed{0} \]