Let ` f(x) = {:{ (x sin""(1/x) , x ne 0) , ( k , x = 0):}`
then f(x) is continuous at x = 0 if

Options

A. k =1
B. k =0
C. k = 2
D. k = -1

Answer

To determine the value of \( k \) for which the function

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

is continuous at \( x = 0 \), we need to check the condition for continuity, which states that:

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

This means we need to find the limit of \( f(x) \) as \( x \) approaches \( 0 \) and set it equal to \( k \).

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

We need to evaluate:

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

### 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 \( x \) (noting that \( x \) approaches \( 0 \) from both sides, so we consider the absolute value):

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

### Step 3: Evaluate the limits of the bounding functions

Now, we take the limit of the bounding functions as \( x \) approaches \( 0 \):

\[ \lim_{x \to 0} -x = 0 \]
\[ \lim_{x \to 0} x = 0 \]

### Step 4: Apply the Squeeze Theorem

Since:

\[ \lim_{x \to 0} -x \leq \lim_{x \to 0} x \sin\left(\frac{1}{x}\right) \leq \lim_{x \to 0} x \]

and both limits on the left and right are equal to \( 0 \), by the Squeeze Theorem, we have:

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

### Step 5: Set the limit equal to \( k \)

Now, since \( f(0) = k \), we set:

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

### Conclusion

Thus, the value of \( k \) for which \( f(x) \) is continuous at \( x = 0 \) is:

\[ \boxed{0} \]