If `f(x)={{:("x sin"(1)/(x)",",xne0),(k",",x=0):}` is continuous at x=0, then the value of k is

Options

A. 1
B. `-1`
C. `0`
D. `2`

Explanation

To determine the value of \( k \) such that 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 ensure that the limit of \( f(x) \) as \( x \) approaches 0 is equal to \( f(0) \).

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

We need to calculate:

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

### Step 2: Analyze the limit.

As \( x \) approaches 0, \( \sin\left(\frac{1}{x}\right) \) oscillates between -1 and 1. Thus, we can bound \( x \sin\left(\frac{1}{x}\right) \):

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

### Step 3: Apply the Squeeze Theorem.

Since both \( -x \) and \( x \) approach 0 as \( x \) approaches 0, we can apply the Squeeze Theorem:

\[ \lim_{x \to 0} -x = 0 \quad \text{and} \quad \lim_{x \to 0} x = 0 \]

Thus,

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

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

For \( f(x) \) to be continuous at \( x = 0 \), we must have:

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

This means:

\[ 0 = k \]

### Conclusion:

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

\[ \boxed{0} \]

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