If the function f(x) defined by
`f(x)={(x"sin"(1)/(x)",", "for "x ne 0),(k",", "for " x=0):}`
is continuous at x = 0, then k is equal to

Options

A. 0
B. 1
C. `-1`
D. `(1)/(2)`

Explanation

To determine the value of \( k \) such that the function \( f(x) \) is continuous at \( x = 0 \), we need to analyze the function defined as:

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

### Step 1: Understand the condition for continuity
For \( f(x) \) to be continuous at \( x = 0 \), the following condition must hold:

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

This means we need to find \( \lim_{x \to 0} f(x) \) and set it equal to \( k \).

### Step 2: Calculate the limit as \( x \) approaches 0
We need to evaluate:

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

### Step 3: Analyze the limit
As \( x \) approaches 0, \( \frac{1}{x} \) approaches infinity. The sine function oscillates between -1 and 1, so:

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

Multiplying the entire inequality by \( x \) (which approaches 0), we have:

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

### Step 4: Apply the Squeeze Theorem
Since both \( -x \) and \( x \) approach 0 as \( x \) approaches 0, by the Squeeze Theorem, we conclude:

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

### Step 5: Set the limit equal to \( k \)
From the continuity condition, we have:

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

Thus, we can set:

\[ k = 0 \]

### Final Answer
The value of \( k \) that makes the function continuous at \( x = 0 \) is:

\[ \boxed{0} \]