The value of f (0), such that `f( x) =( 1)/(x^(2)) ( 1 -cos( sin x ))` can be made continuous at x=0 , is

Options

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

Answer

To find the value of \( f(0) \) such that the function

\[ f(x) = \frac{1}{x^2} (1 - \cos(\sin x)) \]

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

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

First, we substitute \( x = 0 \) into the function:

\[ f(0) = \frac{1}{0^2} (1 - \cos(\sin(0))) = \frac{1}{0} (1 - \cos(0)) = \frac{1}{0} (1 - 1) = \frac{0}{0} \]

This is an indeterminate form, so we need to find the limit as \( x \) approaches 0.

### Step 2: Find the limit of \( f(x) \) as \( x \to 0 \)

We need to compute:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{1 - \cos(\sin x)}{x^2} \]

### Step 3: Apply L'Hôpital's Rule

Since we have the form \( \frac{0}{0} \), we can apply L'Hôpital's Rule. We differentiate the numerator and the denominator.

1. Differentiate the numerator \( 1 - \cos(\sin x) \):
- The derivative of \( 1 \) is \( 0 \).
- The derivative of \( -\cos(\sin x) \) is \( \sin(\sin x) \cdot \cos x \) (using the chain rule).

2. Differentiate the denominator \( x^2 \):
- The derivative is \( 2x \).

Now we apply L'Hôpital's Rule:

\[ \lim_{x \to 0} \frac{1 - \cos(\sin x)}{x^2} = \lim_{x \to 0} \frac{\sin(\sin x) \cdot \cos x}{2x} \]

### Step 4: Evaluate the new limit

Substituting \( x = 0 \):

\[ \sin(\sin(0)) \cdot \cos(0) = \sin(0) \cdot 1 = 0 \]

So we again have the form \( \frac{0}{0} \). We apply L'Hôpital's Rule again:

1. Differentiate the numerator \( \sin(\sin x) \cdot \cos x \):
- Using the product rule:
- Derivative of \( \sin(\sin x) \) is \( \cos(\sin x) \cdot \cos x \) (chain rule).
- Derivative of \( \cos x \) is \( -\sin x \).

So we have:

\[ \frac{d}{dx}(\sin(\sin x) \cdot \cos x) = \cos(\sin x) \cdot \cos x - \sin(\sin x) \cdot \sin x \]

2. Differentiate the denominator \( 2x \):
- The derivative is \( 2 \).

Now we apply L'Hôpital's Rule again:

\[ \lim_{x \to 0} \frac{\cos(\sin x) \cdot \cos x - \sin(\sin x) \cdot \sin x}{2} \]

### Step 5: Evaluate the limit again

Substituting \( x = 0 \):

\[ \cos(\sin(0)) \cdot \cos(0) - \sin(\sin(0)) \cdot \sin(0) = \cos(0) \cdot 1 - 0 \cdot 0 = 1 - 0 = 1 \]

Thus, we have:

\[ \lim_{x \to 0} f(x) = \frac{1}{2} \]

### Conclusion

To make \( f(x) \) continuous at \( x = 0 \), we need to set:

\[ f(0) = \frac{1}{2} \]