` lim_(x to 0) (x tan 2x -2x tan x)/((1- cos 2x)^(2))` equal

Options

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

Answer

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

### Step 1: Rewrite the limit expression
We start with the original limit:
\[ \lim_{x \to 0} \frac{x \tan 2x - 2x \tan x}{(1 - \cos 2x)^2} \]

### Step 2: Factor out common terms
We can factor out \( x \) from the numerator:
\[ = \lim_{x \to 0} \frac{x (\tan 2x - 2 \tan x)}{(1 - \cos 2x)^2} \]

### Step 3: Use the identity for \( 1 - \cos 2x \)
Using the trigonometric identity \( 1 - \cos 2x = 2 \sin^2 x \), we rewrite the denominator:
\[ = \lim_{x \to 0} \frac{x (\tan 2x - 2 \tan x)}{(2 \sin^2 x)^2} = \lim_{x \to 0} \frac{x (\tan 2x - 2 \tan x)}{4 \sin^4 x} \]

### Step 4: Simplify the limit
Now we need to analyze the expression \( \tan 2x - 2 \tan x \). We can use the Taylor series expansion for small values of \( x \):
- \( \tan x \approx x + \frac{x^3}{3} + O(x^5) \)
- \( \tan 2x \approx 2x + \frac{(2x)^3}{3} + O(x^5) = 2x + \frac{8x^3}{3} + O(x^5) \)

Thus,
\[ \tan 2x - 2 \tan x \approx \left(2x + \frac{8x^3}{3}\right) - 2\left(x + \frac{x^3}{3}\right) = 2x + \frac{8x^3}{3} - 2x - \frac{2x^3}{3} = \frac{6x^3}{3} = 2x^3 \]

### Step 5: Substitute back into the limit
Now substituting back into the limit:
\[ = \lim_{x \to 0} \frac{x (2x^3)}{4 \sin^4 x} = \lim_{x \to 0} \frac{2x^4}{4 \sin^4 x} = \lim_{x \to 0} \frac{x^4}{2 \sin^4 x} \]

### Step 6: Use the limit \( \lim_{x \to 0} \frac{x}{\sin x} = 1 \)
Using the fact that \( \lim_{x \to 0} \frac{x}{\sin x} = 1 \), we have:
\[ \sin^4 x \approx x^4 \text{ as } x \to 0 \]
Thus,
\[ \lim_{x \to 0} \frac{x^4}{2 \sin^4 x} = \lim_{x \to 0} \frac{x^4}{2 (x^4)} = \frac{1}{2} \]

### Final Answer
Therefore, the limit evaluates to:
\[ \lim_{x \to 0} \frac{x \tan 2x - 2x \tan x}{(1 - \cos 2x)^2} = \frac{1}{2} \]