Let `f(x)={((sin pix)/(5x)",",x ne0),(k"," , x =0):}` if f(x) is continuous at x = 0, then k is equal to

Options

A. `(pi)/(5)`
B. `(5)/(pi)`
C. 1
D. 0

Explanation

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

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

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

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

### Step-by-Step Solution:

1. **Identify the limit as \( x \) approaches 0:**
We need to find \( \lim_{x \to 0} f(x) \) when \( x \neq 0 \):
\[ \lim_{x \to 0} \frac{\sin(\pi x)}{5x}. \]

2. **Use the limit property:**
We can simplify the limit using the fact that \( \lim_{x \to 0} \frac{\sin(ax)}{ax} = 1 \) for any constant \( a \). Here, we can rewrite the limit:
\[ \lim_{x \to 0} \frac{\sin(\pi x)}{5x} = \lim_{x \to 0} \left(\frac{\sin(\pi x)}{\pi x} \cdot \frac{\pi}{5}\right). \]

3. **Evaluate the limit:**
Now we can evaluate the limit:
\[ \lim_{x \to 0} \frac{\sin(\pi x)}{\pi x} = 1. \]
Therefore,
\[ \lim_{x \to 0} f(x) = \frac{\pi}{5} \cdot 1 = \frac{\pi}{5}. \]

4. **Set the limit equal to \( f(0) \):**
Since \( f(0) = k \), we set the limit equal to \( k \):
\[ k = \frac{\pi}{5}. \]

5. **Conclusion:**
Thus, the value of \( k \) that makes \( f(x) \) continuous at \( x = 0 \) is:
\[ k = \frac{\pi}{5}. \]