If `f(x)={((sin 5x)/(x^(2)+2x)",", x ne 0),(k+(1)/(2)",", x =0):}` is continuous at x = 0, then the value of k is

Options

A. 1
B. `-2`
C. 2
D. `(1)/(2)`

Explanation

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

\[ f(x) = \begin{cases} \frac{\sin(5x)}{x^2 + 2x} & \text{if } x \neq 0 \\ k + \frac{1}{2} & \text{if } x = 0 \end{cases} \]

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

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

### Step 1: Calculate \( f(0) \)
Since \( f(0) = k + \frac{1}{2} \), we will use this value later.

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

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sin(5x)}{x^2 + 2x}. \]

### Step 3: Simplify the limit
We can factor the denominator:

\[ x^2 + 2x = x(x + 2). \]

Thus, we rewrite the limit:

\[ \lim_{x \to 0} \frac{\sin(5x)}{x(x + 2)}. \]

### Step 4: Use the limit property of sine
We know that

\[ \lim_{x \to 0} \frac{\sin(5x)}{5x} = 1. \]

So we can rewrite our limit:

\[ \lim_{x \to 0} \frac{\sin(5x)}{x(x + 2)} = \lim_{x \to 0} \frac{5}{x + 2} \cdot \frac{\sin(5x)}{5x}. \]

### Step 5: Substitute the limit
Now substituting the limit:

\[ = \lim_{x \to 0} \frac{5}{x + 2} \cdot 1 = \frac{5}{0 + 2} = \frac{5}{2}. \]

### Step 6: Set the limit equal to \( f(0) \)
For continuity at \( x = 0 \):

\[ \lim_{x \to 0} f(x) = f(0) \implies \frac{5}{2} = k + \frac{1}{2}. \]

### Step 7: Solve for \( k \)
Now, we solve for \( k \):

\[ k + \frac{1}{2} = \frac{5}{2} \implies k = \frac{5}{2} - \frac{1}{2} = \frac{4}{2} = 2. \]

Thus, the value of \( k \) is

\[ \boxed{2}. \]