If `f(x)` is continuous at `x=0`, where `f(x)={:{((1-cos x)/(x)", for " x!=0),(k ", for" x = 0 ):}`, then `k=`

Options

A. `1/2`
B. `(-1)/(2)`
C. `1/4`
D. 0

Answer

To determine the value of \( k \) for the function \( f(x) \) defined as:

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

we need to ensure that \( f(x) \) is continuous at \( x = 0 \). For continuity at a point, the following condition must hold:

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

### Step 1: Calculate the limit as \( x \) approaches 0 from the right
We first find the limit of \( f(x) \) as \( x \) approaches 0:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{1 - \cos x}{x} \]

### Step 2: Identify the form of the limit
As \( x \) approaches 0, both the numerator and denominator approach 0, resulting in an indeterminate form \( \frac{0}{0} \).

### Step 3: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply 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 on the right exists. We differentiate the numerator and denominator:

- The derivative of the numerator \( 1 - \cos x \) is \( \sin x \).
- The derivative of the denominator \( x \) is \( 1 \).

Thus, we have:

\[ \lim_{x \to 0} \frac{1 - \cos x}{x} = \lim_{x \to 0} \frac{\sin x}{1} \]

### Step 4: Evaluate the limit
Now we evaluate the limit:

\[ \lim_{x \to 0} \sin x = 0 \]

So, we find:

\[ \lim_{x \to 0} \frac{1 - \cos x}{x} = 0 \]

### Step 5: Set the limit equal to \( k \)
Since \( f(0) = k \), we have:

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

Thus, we can set:

\[ 0 = k \]

### Conclusion
Therefore, the value of \( k \) is:

\[ \boxed{0} \]