If f(x) = `{{:((sin 5x)/(x^(2)+2x), x ne0), (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)`

Answer

To determine the value of \( k \) that makes the function \( f(x) \) continuous at \( x = 0 \), we need to follow these steps:

### Step 1: Understand the condition for continuity
A function is continuous at a point if:
\[ \lim_{x \to a} f(x) = f(a) \]
In this case, we need to check the continuity at \( x = 0 \).

### Step 2: Identify the function
The function is defined as:
\[ f(x) = \begin{cases} \frac{\sin(5x)}{x^2 + 2x} & \text{if } x \neq 0 \\ k + \frac{1}{2} & \text{if } x = 0 \end{cases} \]

### Step 3: 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 4: Simplify the limit
Using the fact that \( \lim_{x \to 0} \frac{\sin(5x)}{5x} = 1 \), we can rewrite the limit:
\[ \lim_{x \to 0} \frac{\sin(5x)}{x^2 + 2x} = \lim_{x \to 0} \frac{\sin(5x)}{5x} \cdot \frac{5x}{x^2 + 2x} \]

### Step 5: Evaluate the limit
Calculating the first part:
\[ \lim_{x \to 0} \frac{\sin(5x)}{5x} = 1 \]
Now, we evaluate the second part:
\[ \lim_{x \to 0} \frac{5x}{x^2 + 2x} = \lim_{x \to 0} \frac{5}{x + 2} = \frac{5}{0 + 2} = \frac{5}{2} \]
Thus, we have:
\[ \lim_{x \to 0} f(x) = 1 \cdot \frac{5}{2} = \frac{5}{2} \]

### Step 6: Set the limit equal to the function value at \( x = 0 \)
For continuity at \( x = 0 \):
\[ \lim_{x \to 0} f(x) = f(0) \]
This gives us:
\[ \frac{5}{2} = k + \frac{1}{2} \]

### Step 7: Solve for \( k \)
Subtract \( \frac{1}{2} \) from both sides:
\[ \frac{5}{2} - \frac{1}{2} = k \]
\[ \frac{4}{2} = k \]
\[ k = 2 \]

### Final Answer
Thus, the value of \( k \) is \( 2 \).

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