`lim_(xtooo) (1-cos^(3)4x)/(x^(2))` is equal to

Options

A. 0
B. 12
C. 24
D. 36

Explanation

To solve the limit \( \lim_{x \to 0} \frac{1 - \cos^3(4x)}{x^2} \), we will follow these steps:

### Step 1: Identify the form of the limit
As \( x \to 0 \), \( \cos(4x) \to 1 \). Therefore, \( \cos^3(4x) \to 1 \) and \( 1 - \cos^3(4x) \to 0 \). The denominator \( x^2 \to 0 \) as well. This gives us the indeterminate form \( \frac{0}{0} \).

**Hint:** Check the behavior of the numerator and denominator as \( x \) approaches 0.

### Step 2: Apply L'Hôpital's Rule
Since we have the indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the denominator separately.

### Step 3: Differentiate the numerator and denominator
- The numerator is \( 1 - \cos^3(4x) \). Using the chain rule:
\[ \frac{d}{dx}(1 - \cos^3(4x)) = -3\cos^2(4x) \cdot (-\sin(4x) \cdot 4) = 12 \cos^2(4x) \sin(4x) \]
- The denominator is \( x^2 \), and its derivative is \( 2x \).

### Step 4: Rewrite the limit using the derivatives
Now we can rewrite the limit:
\[ \lim_{x \to 0} \frac{12 \cos^2(4x) \sin(4x)}{2x} \]

### Step 5: Simplify the limit
This simplifies to:
\[ \lim_{x \to 0} \frac{6 \cos^2(4x) \sin(4x)}{x} \]

### Step 6: Substitute \( 4x \) with \( t \)
Let \( t = 4x \), then as \( x \to 0 \), \( t \to 0 \) and \( x = \frac{t}{4} \). The limit becomes:
\[ \lim_{t \to 0} \frac{6 \cos^2(t) \sin(t)}{\frac{t}{4}} = \lim_{t \to 0} \frac{24 \cos^2(t) \sin(t)}{t} \]

### Step 7: Use the limit property \( \lim_{t \to 0} \frac{\sin(t)}{t} = 1 \)
Now we can separate the limit:
\[ \lim_{t \to 0} 24 \cos^2(t) \cdot \lim_{t \to 0} \frac{\sin(t)}{t} = 24 \cdot 1 \cdot \cos^2(0) = 24 \cdot 1 \cdot 1 = 24 \]

### Final Answer
Thus, the limit is:
\[ \lim_{x \to 0} \frac{1 - \cos^3(4x)}{x^2} = 24 \]

---