If `f(x) = {((sin 2x)/(5x), ",", "when" x != 0),(k, ",", "when" x = 0):}` is continuous at x = 0, then the value of k will be

Options

A. 1
B. `2/5`
C. `-2/5`
D. `5/2`

Answer

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

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

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

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

### Step 1: Find \( f(0) \)

Since \( f(0) = k \), we have:

\[ f(0) = k. \]

### Step 2: Calculate \( \lim_{x \to 0} f(x) \)

For \( x \neq 0 \), we have:

\[ f(x) = \frac{\sin(2x)}{5x}. \]

We need to find:

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

### Step 3: Apply L'Hôpital's Rule

Since both the numerator and denominator approach 0 as \( x \to 0 \), we can apply L'Hôpital's Rule:

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

### Step 4: Evaluate the limit

Now substituting \( x = 0 \):

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

### Step 5: Set the limit equal to \( k \)

For continuity at \( x = 0 \):

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

### Conclusion

Thus, the value of \( k \) is

\[ \boxed{\frac{2}{5}}. \]

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