If f(x) = `{{:((1-cosx )/(x) , x ne 0),( x , x=0):}` is continuous at x =0 then the value of k is

Options

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

Answer

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

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

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:

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

### Step 2: Apply L'Hôpital's Rule.

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

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

Here, we differentiate the numerator and the denominator:

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

### Step 3: Evaluate the limit.

Now we can evaluate the limit:

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

Thus,

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

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

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

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

This means:

\[ 0 = k \]

### Conclusion

Therefore, the value of \( k \) that makes the function continuous at \( x = 0 \) is:

\[ \boxed{0} \]

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