Find the value of k for which the function `f(x)= {((2x+3 sin x)/(3x+2sin x),"when " x ne 0),(4k,"when" x= 0):}` is continuous at x= 0

Answer

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

\[ f(x) = \begin{cases} \frac{2x + 3 \sin x}{3x + 2 \sin x} & \text{when } x \neq 0 \\ 4k & \text{when } x = 0 \end{cases} \]

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

### Step 1: Set up the continuity condition
For \( f(x) \) to be continuous at \( x = 0 \), we need:

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

Given \( f(0) = 4k \), we have:

\[ \lim_{x \to 0} f(x) = 4k \]

### Step 2: Calculate the limit as \( x \) approaches 0
We need to evaluate:

\[ \lim_{x \to 0} \frac{2x + 3 \sin x}{3x + 2 \sin x} \]

### Step 3: Substitute and simplify the limit
We can factor out \( x \) from both the numerator and the denominator:

\[ = \lim_{x \to 0} \frac{x(2 + \frac{3 \sin x}{x})}{x(3 + \frac{2 \sin x}{x})} \]

This simplifies to:

\[ = \lim_{x \to 0} \frac{2 + \frac{3 \sin x}{x}}{3 + \frac{2 \sin x}{x}} \]

### Step 4: Apply the limit
Using the known limit \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \):

\[ = \frac{2 + 3 \cdot 1}{3 + 2 \cdot 1} = \frac{2 + 3}{3 + 2} = \frac{5}{5} = 1 \]

### Step 5: Set the limit equal to \( f(0) \)
Now we have:

\[ 1 = 4k \]

### Step 6: Solve for \( k \)
To find \( k \):

\[ k = \frac{1}{4} \]

### Conclusion
Thus, the value of \( k \) for which the function is continuous at \( x = 0 \) is:

\[ \boxed{\frac{1}{4}} \]