The value of k for which the function defined as `f(x) = {{:((1-coskx)/(x sinx), if x ne 0),(1/2, if x = 0):}` continuous at `x = 0 `

Explanation

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

\[ f(x) = \begin{cases} \frac{1 - \cos(kx)}{x \sin(x)} & \text{if } x \neq 0 \\ \frac{1}{2} & \text{if } x = 0 \end{cases} \]

is continuous at \( x = 0 \), we need to ensure that

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

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

From the definition of the function, we have:

\[ f(0) = \frac{1}{2}. \]

### Step 2: Calculate the limit as \( x \) approaches 0

Next, we need to find:

\[ \lim_{x \to 0} \frac{1 - \cos(kx)}{x \sin(x)}. \]

As \( x \) approaches 0, both the numerator and denominator approach 0, resulting in the indeterminate form \( \frac{0}{0} \). Therefore, we can apply L'Hôpital's Rule.

### Step 3: Apply L'Hôpital's Rule

Using L'Hôpital's Rule, we differentiate the numerator and the denominator:

1. Differentiate the numerator \( 1 - \cos(kx) \):
\[ \frac{d}{dx}(1 - \cos(kx)) = k \sin(kx). \]

2. Differentiate the denominator \( x \sin(x) \):
\[ \frac{d}{dx}(x \sin(x)) = \sin(x) + x \cos(x). \]

Now we can rewrite the limit:

\[ \lim_{x \to 0} \frac{k \sin(kx)}{\sin(x) + x \cos(x)}. \]

### Step 4: Evaluate the new limit

As \( x \to 0 \), both \( \sin(kx) \) and \( \sin(x) + x \cos(x) \) again approach 0, leading to another \( \frac{0}{0} \) form. We apply L'Hôpital's Rule again.

1. Differentiate the numerator \( k \sin(kx) \):
\[ \frac{d}{dx}(k \sin(kx)) = k^2 \cos(kx). \]

2. Differentiate the denominator \( \sin(x) + x \cos(x) \):
\[ \frac{d}{dx}(\sin(x) + x \cos(x)) = \cos(x) + \cos(x) - x \sin(x) = 2\cos(x) - x \sin(x). \]

Now we have:

\[ \lim_{x \to 0} \frac{k^2 \cos(kx)}{2\cos(x) - x \sin(x)}. \]

### Step 5: Evaluate the limit as \( x \) approaches 0

Substituting \( x = 0 \):

\[ \lim_{x \to 0} \frac{k^2 \cos(k \cdot 0)}{2\cos(0) - 0 \cdot \sin(0)} = \frac{k^2 \cdot 1}{2 \cdot 1 - 0} = \frac{k^2}{2}. \]

### Step 6: Set the limit equal to \( f(0) \)

For continuity at \( x = 0 \), we set the limit equal to \( f(0) \):

\[ \frac{k^2}{2} = \frac{1}{2}. \]

### Step 7: Solve for \( k \)

Multiplying both sides by 2:

\[ k^2 = 1. \]

Taking the square root:

\[ k = 1 \quad \text{or} \quad k = -1. \]

Thus, the values of \( k \) for which the function is continuous at \( x = 0 \) are:

\[ k = 1 \quad \text{or} \quad k = -1. \]