Find the value of constant k when the function is continuous at x=0, where `f(x)={((1-cos4x)/(8x^2),xne0),(k,x=0):}`

Options

A. 0
B. `-1`
C. 1
D. `1/2`

Answer

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

\[ f(x) = \begin{cases} \frac{1 - \cos(4x)}{8x^2} & \text{if } x \neq 0 \\ k & \text{if } x = 0 \end{cases} \]

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

### Step-by-Step Solution:

1. **Identify the limit as \( x \) approaches 0**:
We need to compute:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{1 - \cos(4x)}{8x^2} \]

2. **Check the form of the limit**:
As \( x \to 0 \), both the numerator and denominator approach 0, resulting in the indeterminate form \( \frac{0}{0} \).

3. **Apply L'Hôpital's Rule**:
Since we have an indeterminate form, we can use L'Hôpital's Rule, which states that:
\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \]
if the limit exists. Here, let \( f(x) = 1 - \cos(4x) \) and \( g(x) = 8x^2 \).

4. **Differentiate the numerator and denominator**:
- The derivative of the numerator \( f(x) = 1 - \cos(4x) \) is:
\[ f'(x) = 4\sin(4x) \]
- The derivative of the denominator \( g(x) = 8x^2 \) is:
\[ g'(x) = 16x \]

5. **Rewrite the limit**:
Now we can rewrite the limit using the derivatives:
\[ \lim_{x \to 0} \frac{1 - \cos(4x)}{8x^2} = \lim_{x \to 0} \frac{4\sin(4x)}{16x} = \lim_{x \to 0} \frac{\sin(4x)}{4x} \]

6. **Evaluate the limit**:
We know that:
\[ \lim_{u \to 0} \frac{\sin(u)}{u} = 1 \]
Therefore,
\[ \lim_{x \to 0} \frac{\sin(4x)}{4x} = 1 \]

7. **Final limit calculation**:
Thus, we have:
\[ \lim_{x \to 0} \frac{1 - \cos(4x)}{8x^2} = 1 \]

8. **Set the limit equal to \( k \)**:
For the function to be continuous at \( x = 0 \):
\[ k = \lim_{x \to 0} f(x) = 1 \]

### Conclusion:
The value of the constant \( k \) is \( 1 \).