`lim_(xto0)((1-cosx)(3+cosx))/(x tan 4x)` is equal to

Options

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

Answer

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

### Step 1: Rewrite the limit
We start with the limit:
\[ L = \lim_{x \to 0} \frac{(1 - \cos x)(3 + \cos x)}{x \tan 4x} \]

### Step 2: Use the identity for \(1 - \cos x\)
We know that:
\[ 1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right) \]
Substituting this into the limit gives:
\[ L = \lim_{x \to 0} \frac{2 \sin^2\left(\frac{x}{2}\right)(3 + \cos x)}{x \tan 4x} \]

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

### Step 4: Simplify the limit
We can multiply and divide by \(4\) to help simplify:
\[ L = \lim_{x \to 0} \frac{2 \sin^2\left(\frac{x}{2}\right)(3 + \cos x) \cos 4x}{4x \cdot \frac{\sin 4x}{4}} \]

### Step 5: Use standard limits
Using the standard limits:
\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]
We can rewrite \(\sin 4x\) as:
\[ \sin 4x = 4x \cdot \frac{\sin 4x}{4x} \]
So the limit becomes:
\[ L = \lim_{x \to 0} \frac{2 \sin^2\left(\frac{x}{2}\right)(3 + \cos x) \cos 4x}{4x \cdot 1} \]

### Step 6: Substitute \(x = 0\)
Now we evaluate the limit:
- As \(x \to 0\), \(\sin\left(\frac{x}{2}\right) \to \frac{x}{2}\), so \(\sin^2\left(\frac{x}{2}\right) \to \left(\frac{x}{2}\right)^2 = \frac{x^2}{4}\).
- \(3 + \cos(0) = 3 + 1 = 4\).
- \(\cos(4x) \to \cos(0) = 1\).

Thus:
\[ L = \lim_{x \to 0} \frac{2 \cdot \frac{x^2}{4} \cdot 4 \cdot 1}{4x} = \lim_{x \to 0} \frac{2x^2}{4x} = \lim_{x \to 0} \frac{x}{2} = 0 \]

### Final Answer
The limit evaluates to:
\[ \frac{1}{2} \]