$$\lim_{x \to 0} \frac{x\tan 2x - 2x\tan x}{(1 - \cos 2x)^2}$$ equals:

We are asked to evaluate the limit

$$\lim_{x \to 0}\dfrac{x\tan 2x\;-\;2x\tan x}{\left(1-\cos 2x\right)^{2}}.$$

Because both numerator and denominator tend to $$0$$ as $$x\to 0$$, this is an indeterminate form $$\dfrac{0}{0}$$. In such cases the Maclaurin (Taylor) expansions of the standard trigonometric functions about $$x=0$$ are very useful. We first recall the series we will need:

For small $$x$$,

$$\tan x = x + \dfrac{x^{3}}{3} + \dfrac{2x^{5}}{15} + \cdots,$$

$$\cos x = 1 - \dfrac{x^{2}}{2} + \dfrac{x^{4}}{24} - \dfrac{x^{6}}{720} + \cdots.$$

Now we substitute $$x$$ by $$2x$$ wherever required and write every term up to the power that will finally survive after simplification.

1. Expansion of $$\tan 2x$$

Replacing $$x$$ with $$2x$$ in the series of $$\tan x$$ we get

$$\tan 2x = 2x + \dfrac{(2x)^{3}}{3} + \dfrac{2(2x)^{5}}{15} + \cdots = 2x + \dfrac{8x^{3}}{3} + \dfrac{32x^{5}}{15} + \cdots.$$

2. Expansion of $$x\tan 2x$$

Multiplying by the outside factor $$x$$ gives

$$x\tan 2x = x\left(2x + \dfrac{8x^{3}}{3} + \dfrac{32x^{5}}{15} + \cdots\right) = 2x^{2} + \dfrac{8x^{4}}{3} + \dfrac{32x^{6}}{15} + \cdots.$$

3. Expansion of $$2x\tan x$$

First write $$\tan x = x + \dfrac{x^{3}}{3} + \dfrac{2x^{5}}{15} + \cdots,$$ then multiply by $$2x$$:

$$2x\tan x = 2x\left(x + \dfrac{x^{3}}{3} + \dfrac{2x^{5}}{15} + \cdots\right) = 2x^{2} + \dfrac{2x^{4}}{3} + \dfrac{4x^{6}}{15} + \cdots.$$

4. Numerator $$x\tan 2x - 2x\tan x$$

Subtracting series term-by-term we obtain

$$x\tan 2x - 2x\tan x = \bigl(2x^{2} + \tfrac{8x^{4}}{3} + \tfrac{32x^{6}}{15} + \cdots\bigr) \;-\;\bigl(2x^{2} + \tfrac{2x^{4}}{3} + \tfrac{4x^{6}}{15} + \cdots\bigr).$$

The $$2x^{2}$$ terms cancel, giving

$$x\tan 2x - 2x\tan x = \left(\dfrac{8x^{4}}{3}-\dfrac{2x^{4}}{3}\right) + \left(\dfrac{32x^{6}}{15}-\dfrac{4x^{6}}{15}\right) + \cdots = 2x^{4} + \dfrac{28x^{6}}{15} + \cdots.$$

Thus, up to the smallest non-zero power, the numerator is

$$x\tan 2x - 2x\tan x = 2x^{4} + O(x^{6}).$$

5. Expansion of $$1-\cos 2x$$

Substituting $$x$$ by $$2x$$ in the cosine series gives

$$\cos 2x = 1 - \dfrac{(2x)^{2}}{2} + \dfrac{(2x)^{4}}{24} - \dfrac{(2x)^{6}}{720} + \cdots = 1 - 2x^{2} + \dfrac{2x^{4}}{3} - \dfrac{4x^{6}}{45} + \cdots.$$

Therefore

$$1 - \cos 2x = 2x^{2} - \dfrac{2x^{4}}{3} + \dfrac{4x^{6}}{45} + \cdots.$$

6. Denominator $$\left(1-\cos 2x\right)^{2}$$

We square the above expression, keeping terms up to $$x^{6}$$ (because the numerator starts from $$x^{4}$$):

$$\left(1-\cos 2x\right)^{2} = \left(2x^{2} - \dfrac{2x^{4}}{3} + \cdots\right)^{2}.$$

Using the algebraic identity $$(a+b)^{2}=a^{2}+2ab+b^{2},$$ and remembering that any term of power higher than $$x^{6}$$ can be ignored for the limit, we compute:

First term: $$\bigl(2x^{2}\bigr)^{2}=4x^{4}.$$
Cross term: $$2\left(2x^{2}\right)\left(-\dfrac{2x^{4}}{3}\right)=-\dfrac{8x^{6}}{3}.$$
Square of the smaller term $$\left(-\dfrac{2x^{4}}{3}\right)^{2}$$ is of order $$x^{8}$$ and can be dropped.

Thus

$$\left(1-\cos 2x\right)^{2} = 4x^{4} - \dfrac{8x^{6}}{3} + \cdots = 4x^{4} + O(x^{6}).$$

7. Forming the ratio

We now divide the expanded numerator by the expanded denominator:

$$\dfrac{x\tan 2x - 2x\tan x}{\left(1-\cos 2x\right)^{2}} = \dfrac{\,2x^{4} + \dfrac{28x^{6}}{15} + \cdots\,} {\,4x^{4} - \dfrac{8x^{6}}{3} + \cdots\,}.$$

Factor $$x^{4}$$ out of both numerator and denominator and cancel it:

$$= \dfrac{\,2 + \dfrac{28x^{2}}{15} + \cdots\,} {\,4 - \dfrac{8x^{2}}{3} + \cdots\,}.$$

Now let $$x\to 0$$. All the terms containing positive powers of $$x$$ vanish, leaving

$$\lim_{x \to 0}\dfrac{x\tan 2x - 2x\tan x}{(1-\cos 2x)^{2}} = \dfrac{2}{4} = \dfrac{1}{2}.$$

Hence, the correct answer is Option D.