$$\lim_{x \to 0} \frac{x \cot(4x)}{\sin^2 x \cot^2(2x)}$$ is equal to:

We start with the given limit

$$L=\lim_{x\to 0}\frac{x\cot(4x)}{\sin^{2}x\;\cot^{2}(2x)}.$$

First we express every $$\cot$$ in terms of $$\sin$$ and $$\cos$$:

$$\cot(4x)=\frac{\cos(4x)}{\sin(4x)},\qquad \cot(2x)=\frac{\cos(2x)}{\sin(2x)}.$$

Substituting these in $$L$$ we obtain

$$L=\lim_{x\to 0}\frac{x\displaystyle\frac{\cos(4x)}{\sin(4x)}}{\sin^{2}x \left(\displaystyle\frac{\cos(2x)}{\sin(2x)}\right)^{2}} =\lim_{x\to 0}\frac{x\cos(4x)}{\sin(4x)} \;\frac{\sin^{2}(2x)}{\sin^{2}x\,\cos^{2}(2x)}.$$

Now we use the double-angle identity $$\sin(2x)=2\sin x\cos x$$, so

$$\sin^{2}(2x)=4\sin^{2}x\cos^{2}x.$$

Substituting this form of $$\sin^{2}(2x)$$ gives

$$L=\lim_{x\to 0}\frac{x\cos(4x)}{\sin(4x)} \;\frac{4\sin^{2}x\cos^{2}x}{\sin^{2}x\cos^{2}(2x)} =\lim_{x\to 0}\frac{4x\cos(4x)\cos^{2}x}{\sin(4x)\cos^{2}(2x)}.$$

The $$\sin^{2}x$$ terms have cancelled, leaving

$$L=\lim_{x\to 0}\Bigl(\frac{4x}{\sin(4x)}\Bigr)\; \Bigl(\cos(4x)\Bigr)\; \Bigl(\frac{\cos^{2}x}{\cos^{2}(2x)}\Bigr).$$

We can now evaluate each factor separately by using the standard limits

$$\lim_{t\to 0}\frac{\sin t}{t}=1\quad\text{and}\quad \lim_{t\to 0}\cos t=1.$$

For the first factor we put $$t=4x$$, so when $$x\to 0$$, $$t\to 0$$ and

$$\frac{4x}{\sin(4x)}=\frac{t}{\sin t}\longrightarrow 1.$$

The second factor tends to

$$\cos(4x)\longrightarrow 1.$$

For the third factor we use $$\cos x\longrightarrow 1$$ and $$\cos(2x)\longrightarrow 1$$, hence

$$\frac{\cos^{2}x}{\cos^{2}(2x)}\longrightarrow \frac{1^{2}}{1^{2}}=1.$$

Multiplying these three limiting values we get

$$L=1\times 1\times 1=1.$$

Hence, the correct answer is Option D.