1. We need to evaluate $L=\lim_{x\to 0}\frac{x\cot(4x)}{\sin^2 x\,\cot^2(2x)}.$

2. Rewrite cotangent in terms of sine and cosine: $\cot(4x)=\frac{\cos 4x}{\sin 4x}, \qquad \cot(2x)=\frac{\cos 2x}{\sin 2x}.$ So $L=\lim_{x\to 0}\frac{x\,\dfrac{\cos 4x}{\sin 4x}}{\sin^2 x\,\dfrac{\cos^2 2x}{\sin^2 2x}}.$

3. Simplify: $L=\lim_{x\to 0}x\cdot \frac{\cos 4x}{\sin 4x}\cdot \frac{\sin^2 2x}{\sin^2 x\cos^2 2x}.$ Using $\sin 2x=2\sin x\cos x,$ we get $\sin^2 2x=4\sin^2 x\cos^2 x.$ Therefore, $L=\lim_{x\to 0}x\cdot \frac{\cos 4x}{\sin 4x}\cdot \frac{4\sin^2 x\cos^2 x}{\sin^2 x\cos^2 2x}$ $=\lim_{x\to 0}4x\cdot \frac{\cos 4x}{\sin 4x}\cdot \frac{\cos^2 x}{\cos^2 2x}.$

4. Now use the standard limit $\lim_{t\to 0}\frac{\sin t}{t}=1 \quad \Rightarrow \quad \lim_{t\to 0}\frac{t}{\sin t}=1.$ Hence, $\lim_{x\to 0}\frac{4x}{\sin 4x}=1.$ Also, $\cos 4x\to 1, \qquad \cos^2 x\to 1, \qquad \cos^2 2x\to 1 \quad (x\to 0).$ So, $L=1\cdot 1\cdot 1=1.$

5. Therefore the value of the limit is $\boxed{1}.$

6. Checking options:

• A: $0$ ❌
• B: $4$ ❌
• C: $1$ ✅
• D: $2$ ❌

Hence, the correct option is C.