mathop {lim }limits_{x to 0} ,frac{{x,cot ,le | vedclass

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(D) Given limit: $L = \mathop {\lim }\limits_{x 	o 0} \frac{x \cot(4x)}{\sin^2 x \cot^2(2x)}$Substitute $\cot 	heta = \frac{\cos 	heta}{\sin 	heta

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(D) Given limit: $L = \mathop {\lim }\limits_{x 	o 0} \frac{x \cot(4x)}{\sin^2 x \cot^2(2x)}$Substitute $\cot 	heta = \frac{\cos 	heta}{\sin 	heta}$:$L = \mathop {\lim }\limits_{x 	o 0} \frac{x \cos(4x) \sin^2(2x)}{\sin^2 x \sin(4x) \cos^2(2x)}$Use the identity $\sin(4x) = 2 \sin(2x) \cos(2x)$:$L = \mathop {\lim }\limits_{x 	o 0} \frac{x \cos(4x) \sin^2(2x)}{\sin^2 x (2 \sin(2x) \cos(2x)) \cos^2(2x)}$Simplify the expression:$L = \mathop {\lim }\limits_{x 	o 0} \left( \frac{x}{\sin x} ight)^2 \cdot \frac{\sin(2x)}{2x} \cdot \frac{\cos(4x)}{\cos^3(2x)}$As $x 	o 0$,$\frac{\sin 	heta}{	heta} 	o 1$ and $\cos 	heta 	o 1$:$L = (1)^2 \cdot 1 \cdot \frac{1}{1^3} = 1$

$\mathop {\lim }\limits_{x \to 0} \,\frac{{x\,\cot \,\left( {4x} \right)}}{{{{\sin }^2}\,x\,{{\cot }^2}\,\left( {2x} \right)}}$ is equal to

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