Find the value of k for which the function.
` f (x) = {{:(((sin x +xcos x))/( x ) , " when " x ne 0 ),( k, " when" x=0 ):} " is continuous at "x=0 `

Answer

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

\[ f(x) = \begin{cases} \frac{\sin x + x \cos x}{x} & \text{when } x \neq 0 \\ k & \text{when } x = 0 \end{cases} \]

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 compute:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sin x + x \cos x}{x} \]

### Step 2: Split the limit into two parts.

We can separate the limit into two parts:

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

This simplifies to:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sin x}{x} + \lim_{x \to 0} \cos x \]

### Step 3: Evaluate the limits.

From standard limit results, we know:

\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]

And,

\[ \lim_{x \to 0} \cos x = \cos(0) = 1 \]

### Step 4: Combine the results.

Now, we combine the results:

\[ \lim_{x \to 0} f(x) = 1 + 1 = 2 \]

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

For the function to be continuous at \( x = 0 \), we set:

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

Thus, we have:

\[ k = 2 \]

### Conclusion:

The value of \( k \) for which the function is continuous at \( x = 0 \) is

\[ \boxed{2} \]

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