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

We have to evaluate the limit

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

Whenever a limit involves trigonometric functions at the point $$x\to 0$$, the standard small-angle expansions are extremely helpful. Yet, before jumping to series, we can simplify the given expression algebraically with well-known identities.

First, recall the double-angle identity for cosine:

$$\cos 2x = 1 - 2\sin^{2}x.$$

From this, we obtain

$$1-\cos 2x = 1-\bigl(1-2\sin^{2}x\bigr)=2\sin^{2}x.$$

Substituting this into the numerator, we get

$$(1-\cos 2x)^{2}=\bigl(2\sin^{2}x\bigr)^{2}=4\sin^{4}x.$$

Next, let us turn to the denominator. Factor out the common factor $$x:$$

$$2x\tan x - x\tan 2x = x\bigl(2\tan x - \tan 2x\bigr).$$

At this stage, the entire limit becomes

$$\lim_{x\to 0}\;\frac{4\sin^{4}x}{x\bigl(2\tan x - \tan 2x\bigr)}.$$

Now we introduce the standard Maclaurin (small-angle) expansions up to the terms required for accuracy:

• For sine: $$\sin x = x - \dfrac{x^{3}}{6} + O(x^{5}).$$
• For tangent: $$\tan x = x + \dfrac{x^{3}}{3} + O(x^{5}).$$
• Consequently, $$\tan 2x = 2x + \dfrac{(2x)^{3}}{3} + O(x^{5}) = 2x + \dfrac{8x^{3}}{3} + O(x^{5}).$$

Let us treat the numerator first. Using $$\sin x = x - \dfrac{x^{3}}{6}+O(x^{5})$$ we find

$$\sin^{2}x \;=\;\bigl(x - \tfrac{x^{3}}{6}\bigr)^{2} = x^{2} - \tfrac{x^{4}}{3} + O(x^{6}).$$

Therefore,

$$\sin^{4}x = \bigl(\sin^{2}x\bigr)^{2} = \bigl(x^{2} - \tfrac{x^{4}}{3}\bigr)^{2} = x^{4} - \tfrac{2x^{6}}{3} + O(x^{8}).$$

Multiplying by 4 (from the earlier factor) gives

$$4\sin^{4}x = 4x^{4} - \tfrac{8x^{6}}{3} + O(x^{8}).$$

So, up to the order we need, the numerator behaves like $$4x^{4}+O(x^{6}).$$

Now handle the bracket in the denominator:

$$$ 2\tan x - \tan 2x = 2\Bigl(x + \tfrac{x^{3}}{3}\Bigr) - \Bigl(2x + \tfrac{8x^{3}}{3}\Bigr) + O(x^{5}). $$$

Simplifying term by term,

$$$ 2x + \tfrac{2x^{3}}{3} - 2x - \tfrac{8x^{3}}{3} = -\tfrac{6x^{3}}{3} = -2x^{3} + O(x^{5}). $$$

Multiplying by the external factor $$x$$ (which we had factored earlier) gives the entire denominator:

$$$ x\bigl(2\tan x - \tan 2x\bigr) = x\bigl(-2x^{3} + O(x^{5})\bigr) = -2x^{4} + O(x^{6}). $$$

Now substitute the approximations of numerator and denominator back into the limit expression:

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

Since both numerator and denominator share the common factor $$x^{4}$$, we may cancel it, leaving

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

As $$x \to 0$$, all the higher-order terms $$O(x^{2})$$ vanish, giving

$$\lim_{x\to 0}\;\frac{4}{-2} \;=\; -2.$$

Hence, the required limit equals $$-2$$.

Hence, the correct answer is Option C.