Evaluate `lim_(xto0) (xtan2x-2xtanx)/((1-cos2x)^(2)).`

Explanation

To evaluate the limit

\[ \lim_{x \to 0} \frac{x \tan(2x) - 2x \tan(x)}{(1 - \cos(2x))^2}, \]

we will follow these steps:

### Step 1: Rewrite the expression
We start with the limit expression:

\[ \lim_{x \to 0} \frac{x \tan(2x) - 2x \tan(x)}{(1 - \cos(2x))^2}. \]

### Step 2: Use the identity for \( \cos(2x) \)
We know that \( \cos(2x) = 1 - 2\sin^2(x) \). Therefore,

\[ 1 - \cos(2x) = 2\sin^2(x). \]

Substituting this into the limit gives:

\[ \lim_{x \to 0} \frac{x \tan(2x) - 2x \tan(x)}{(2\sin^2(x))^2} = \lim_{x \to 0} \frac{x \tan(2x) - 2x \tan(x)}{4\sin^4(x)}. \]

### Step 3: Factor out \( x \) from the numerator
Now, we can factor \( x \) out of the numerator:

\[ = \lim_{x \to 0} \frac{x(\tan(2x) - 2\tan(x))}{4\sin^4(x)}. \]

### Step 4: Simplify the limit
We can rewrite the limit as:

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

### Step 5: Use Taylor series expansion
Using the Taylor series expansion for small \( x \):

\[ \tan(x) \approx x + \frac{x^3}{3} + O(x^5), \]

we have:

\[ \tan(2x) \approx 2x + \frac{(2x)^3}{3} = 2x + \frac{8x^3}{3}. \]

Thus,

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

### Step 6: Substitute back into the limit
Now substituting back into the limit gives:

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

### Step 7: Use the small angle approximation for \( \sin(x) \)
For small \( x \), \( \sin(x) \approx x \). Therefore, \( \sin^4(x) \approx x^4 \):

\[ \lim_{x \to 0} \frac{2x^3}{4x^4} = \lim_{x \to 0} \frac{2}{4x} = \lim_{x \to 0} \frac{1}{2x}. \]

### Step 8: Evaluate the limit
As \( x \to 0 \), \( \frac{1}{2x} \to \infty \). Thus, the limit diverges.

### Final Result
The limit diverges to \( \infty \).