The value of k for which `f(x) = {((sin 5x)/(3x)","," if " x !=0),(" k,"," if " x = 0):}` is contnuous at x = 0 is

Options

A. `(1)/(3)`
B. 0
C. `(3)/(5)`
D. `(5)/(3)`

Explanation

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

\[ f(x) = \begin{cases} \frac{\sin(5x)}{3x} & \text{if } x \neq 0 \\ k & \text{if } x = 0 \end{cases} \]

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

### Step 1: Find the limit of \( f(x) \) as \( x \) approaches 0.

We need to calculate:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sin(5x)}{3x} \]

### Step 2: Use the limit property of sine.

We know from calculus that:

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

To apply this property, we can rewrite the limit:

\[ \lim_{x \to 0} \frac{\sin(5x)}{3x} = \lim_{x \to 0} \frac{\sin(5x)}{5x} \cdot \frac{5}{3} \]

### Step 3: Evaluate the limit.

Now, we can evaluate the limit:

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

Thus, we have:

\[ \lim_{x \to 0} f(x) = \frac{5}{3} \cdot 1 = \frac{5}{3} \]

### Step 4: Set the limit equal to \( f(0) \).

For the function to be continuous at \( x = 0 \), we need:

\[ f(0) = k \]

So we set:

\[ k = \frac{5}{3} \]

### Conclusion

The value of \( k \) for which \( f(x) \) is continuous at \( x = 0 \) is:

\[ \boxed{\frac{5}{3}} \]

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