1. We need to evaluate

$$L=\lim_{x\to 0}\frac{x\tan 2x-2x\tan x}{(1-\cos 2x)^2}.$$

2. Use standard small-angle expansions near $x=0$:

$$\tan x = x+\frac{x^3}{3}+O(x^5),$$

so

$$\tan 2x = 2x+\frac{(2x)^3}{3}+O(x^5)=2x+\frac{8x^3}{3}+O(x^5).$$

Thus,

$$x\tan 2x = x\left(2x+\frac{8x^3}{3}+O(x^5)\right)=2x^2+\frac{8x^4}{3}+O(x^6),$$

and

$$2x\tan x = 2x\left(x+\frac{x^3}{3}+O(x^5)\right)=2x^2+\frac{2x^4}{3}+O(x^6).$$

Therefore the numerator is

$$x\tan 2x-2x\tan x =\left(2x^2+\frac{8x^4}{3}\right)-\left(2x^2+\frac{2x^4}{3}\right)+O(x^6) =2x^4+O(x^6).$$

3. Now expand the denominator:

$$\cos 2x = 1-\frac{(2x)^2}{2}+O(x^4)=1-2x^2+O(x^4).$$

Hence,

$$1-\cos 2x = 2x^2+O(x^4).$$

So

$$(1-\cos 2x)^2 = (2x^2+O(x^4))^2 = 4x^4+O(x^6).$$

4. Therefore,

$$L=\lim_{x\to 0}\frac{2x^4+O(x^6)}{4x^4+O(x^6)}=\frac{2}{4}=\frac12.$$

5. Check options:

• A: $\frac14$
• B: $1$
• C: $\frac12$ ✅
• D: $-\frac12$

So the correct option is C.