`lim_(x to 0) ((1-cos 2x)(3+cosx))/(x tan 4x)` is equal to

Options

A. 4
B. 3
C. 2
D. `(1)/(2)`

Explanation

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

### Step 1: Use the identity for \( \cos 2x \)
We know that:
\[ \cos 2x = 1 - 2 \sin^2 x \]
Thus, we can rewrite \( 1 - \cos 2x \) as:
\[ 1 - \cos 2x = 2 \sin^2 x \]

### Step 2: Substitute in the limit
Substituting this into the limit, we have:
\[ \lim_{x \to 0} \frac{(2 \sin^2 x)(3 + \cos x)}{x \tan 4x} \]

### Step 3: Rewrite \( \tan 4x \)
Recall that \( \tan 4x = \frac{\sin 4x}{\cos 4x} \). Therefore, we can rewrite the limit as:
\[ \lim_{x \to 0} \frac{(2 \sin^2 x)(3 + \cos x)}{x \cdot \frac{\sin 4x}{\cos 4x}} = \lim_{x \to 0} \frac{(2 \sin^2 x)(3 + \cos x) \cos 4x}{x \sin 4x} \]

### Step 4: Break the limit into parts
We can break this limit into two parts:
\[ \lim_{x \to 0} \frac{2 \sin^2 x}{x} \cdot \lim_{x \to 0} (3 + \cos x) \cdot \lim_{x \to 0} \frac{\cos 4x}{\sin 4x} \]

### Step 5: Evaluate each limit
1. **First Limit**:
\[ \lim_{x \to 0} \frac{2 \sin^2 x}{x} = \lim_{x \to 0} 2 \cdot \frac{\sin^2 x}{x^2} \cdot x = 2 \cdot 1 \cdot 0 = 0 \]
(Note: This limit is actually \( 0 \) since \( \sin x \approx x \) as \( x \to 0 \))

2. **Second Limit**:
\[ \lim_{x \to 0} (3 + \cos x) = 3 + 1 = 4 \]

3. **Third Limit**:
\[ \lim_{x \to 0} \frac{\cos 4x}{\sin 4x} = \lim_{x \to 0} \frac{1}{4} \cdot \frac{\cos 4x}{x} = \frac{1}{4} \cdot 1 = \frac{1}{4} \]

### Step 6: Combine the results
Now, combining these results:
\[ \lim_{x \to 0} \frac{(1 - \cos 2x)(3 + \cos x)}{x \tan 4x} = 0 \cdot 4 \cdot \frac{1}{4} = 0 \]

### Final Result:
Thus, the limit is:
\[ \boxed{0} \]