For what values of k , `f(x)={:{((1-coskx)/(xsinx)," if " x ne 0),(1/2," if " x =0):}` is continuous at x = 0 ?

Answer

To determine the values 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

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

Given that \( f(0) = \frac{1}{2} \), we need to compute the limit:

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

### Step 1: Compute the limit

Using the fact that \( \sin(x) \) approaches \( x \) as \( x \) approaches \( 0 \), we can rewrite \( \sin(x) \) in the limit:

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

### Step 2: Apply the limit properties

We know that

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

Thus, we can simplify the limit to:

\[ \lim_{x \to 0} \frac{1 - \cos(kx)}{x^2}. \]

### Step 3: Use the trigonometric identity

Using the identity \( 1 - \cos(kx) = \sin^2\left(\frac{kx}{2}\right) \) gives us:

\[ \lim_{x \to 0} \frac{\sin^2\left(\frac{kx}{2}\right)}{x^2}. \]

### Step 4: Change of variable

Let \( u = \frac{kx}{2} \), then as \( x \to 0 \), \( u \to 0 \) as well. Therefore, we can rewrite \( x \) in terms of \( u \):

\[ x = \frac{2u}{k}. \]

Substituting this into the limit gives:

\[ \lim_{u \to 0} \frac{\sin^2(u)}{\left(\frac{2u}{k}\right)^2} = \lim_{u \to 0} \frac{k^2 \sin^2(u)}{4u^2}. \]

### Step 5: Evaluate the limit

Using the fact that \( \lim_{u \to 0} \frac{\sin(u)}{u} = 1 \):

\[ \lim_{u \to 0} \frac{\sin^2(u)}{u^2} = 1. \]

Thus, we have:

\[ \lim_{u \to 0} \frac{k^2 \sin^2(u)}{4u^2} = \frac{k^2}{4}. \]

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

Now we set this equal to \( f(0) \):

\[ \frac{k^2}{4} = \frac{1}{2}. \]

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

Multiplying both sides by 4 gives:

\[ k^2 = 2. \]

Taking the square root of both sides results in:

\[ k = \pm \sqrt{2}. \]

### Conclusion

The values of \( k \) for which \( f(x) \) is continuous at \( x = 0 \) are:

\[ k = \sqrt{2} \quad \text{and} \quad k = -\sqrt{2}. \]