`f(x) = {{:((1-cosAx)/(x sinx), if x ne 0),(1/2, if x = 0):}` at `x = 0 `

Answer

To determine the value of \( A \) such that the function

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

is continuous at \( x = 0 \), we need to ensure that the limit of \( f(x) \) as \( x \) approaches 0 equals \( f(0) \).

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

From the definition of the function, we have:

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

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

We need to compute:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{1 - \cos(Ax)}{x \sin x} \]

### Step 3: Use the identity \( 1 - \cos(Ax) = 2 \sin^2\left(\frac{Ax}{2}\right) \)

Substituting this identity into the limit gives:

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

### Step 4: Rewrite the limit

We can rewrite the limit as:

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

### Step 5: Apply the limit \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \)

We know that:

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

Thus, we can rewrite the limit as:

\[ \lim_{x \to 0} \frac{2 \sin^2\left(\frac{Ax}{2}\right)}{x^2} \cdot \frac{1}{\lim_{x \to 0} \frac{\sin x}{x}} = \lim_{x \to 0} \frac{2 \sin^2\left(\frac{Ax}{2}\right)}{x^2} \]

### Step 6: Use the substitution \( u = \frac{Ax}{2} \)

Let \( u = \frac{Ax}{2} \), then as \( x \to 0 \), \( u \to 0 \) as well. Therefore, we have:

\[ x = \frac{2u}{A} \implies x^2 = \frac{4u^2}{A^2} \]

Substituting this into the limit gives:

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

### Step 7: Evaluate the limit

Using \( \lim_{u \to 0} \frac{\sin^2(u)}{u^2} = 1 \):

\[ \lim_{x \to 0} f(x) = \frac{A^2}{2} \cdot 1 = \frac{A^2}{2} \]

### Step 8: Set the limit equal to \( f(0) \)

For continuity at \( x = 0 \):

\[ \frac{A^2}{2} = \frac{1}{2} \]

### Step 9: Solve for \( A \)

Multiplying both sides by 2 gives:

\[ A^2 = 1 \]

Taking the square root of both sides results in:

\[ A = \pm 1 \]

### Final Answer

Thus, the values of \( A \) that make the function continuous at \( x = 0 \) are:

\[ A = 1 \quad \text{or} \quad A = -1 \]