If `f(x)` is continuous at `x=0`, where `f(x)=(3-4cos x+cos 2x)/(x^(4))`, for `x!=0`, then `f(0)=`

Options

A. `1/4`
B. `1/2`
C. 8
D. 4

Answer

To find \( f(0) \) for the function

\[ f(x) = \frac{3 - 4\cos x + \cos 2x}{x^4} \quad \text{for } x \neq 0, \]

we need to ensure that \( f(x) \) is continuous at \( x = 0 \). This means we need to find the limit of \( f(x) \) as \( x \) approaches 0 and set that equal to \( f(0) \).

### Step 1: Find the limit as \( x \to 0 \)

We need to calculate:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{3 - 4\cos x + \cos 2x}{x^4}. \]

### Step 2: Evaluate the limit

As \( x \to 0 \), both the numerator and denominator approach 0, giving us the indeterminate form \( \frac{0}{0} \). We can apply L'Hôpital's Rule, which states that if we have an indeterminate form, we can take the derivative of the numerator and the denominator.

### Step 3: Differentiate the numerator and denominator

1. Differentiate the numerator:
- The derivative of \( 3 \) is \( 0 \).
- The derivative of \( -4\cos x \) is \( 4\sin x \).
- The derivative of \( \cos 2x \) is \( -2\sin 2x \).

Thus, the derivative of the numerator is:

\[ 4\sin x - 2\sin 2x. \]

2. Differentiate the denominator:
- The derivative of \( x^4 \) is \( 4x^3 \).

Now we can rewrite the limit:

\[ \lim_{x \to 0} \frac{4\sin x - 2\sin 2x}{4x^3}. \]

### Step 4: Apply L'Hôpital's Rule again

As \( x \to 0 \), we still have the indeterminate form \( \frac{0}{0} \). We apply L'Hôpital's Rule again:

1. Differentiate the new numerator:
- The derivative of \( 4\sin x \) is \( 4\cos x \).
- The derivative of \( -2\sin 2x \) is \( -4\cos 2x \).

Thus, the new numerator is:

\[ 4\cos x - 4\cos 2x. \]

2. Differentiate the denominator:
- The derivative of \( 4x^3 \) is \( 12x^2 \).

Now we have:

\[ \lim_{x \to 0} \frac{4\cos x - 4\cos 2x}{12x^2}. \]

### Step 5: Apply L'Hôpital's Rule again

We still have the indeterminate form \( \frac{0}{0} \). We apply L'Hôpital's Rule once more:

1. Differentiate the new numerator:
- The derivative of \( 4\cos x \) is \( -4\sin x \).
- The derivative of \( -4\cos 2x \) is \( 8\sin 2x \).

Thus, the new numerator is:

\[ -4\sin x + 8\sin 2x. \]

2. Differentiate the denominator:
- The derivative of \( 12x^2 \) is \( 24x \).

Now we have:

\[ \lim_{x \to 0} \frac{-4\sin x + 8\sin 2x}{24x}. \]

### Step 6: Apply L'Hôpital's Rule again

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

1. Differentiate the new numerator:
- The derivative of \( -4\sin x \) is \( -4\cos x \).
- The derivative of \( 8\sin 2x \) is \( 16\cos 2x \).

Thus, the new numerator is:

\[ -4\cos x + 16\cos 2x. \]

2. Differentiate the denominator:
- The derivative of \( 24x \) is \( 24 \).

Now we have:

\[ \lim_{x \to 0} \frac{-4\cos x + 16\cos 2x}{24}. \]

### Step 7: Evaluate the limit

Now we can substitute \( x = 0 \):

\[ \frac{-4\cos(0) + 16\cos(0)}{24} = \frac{-4(1) + 16(1)}{24} = \frac{12}{24} = \frac{1}{2}. \]

### Conclusion

Since \( f(x) \) is continuous at \( x = 0 \), we have:

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

Thus, the value of \( f(0) \) is

\[ \boxed{\frac{1}{2}}. \]