\mathop {\lim }\limits_{x o 0} {{x\cot \left( {4x} ight)} \over {{{\sin }^2}x{{\cot }^2}\left( {2x} ight)}} is equal to :

Question

\mathop {\lim }\limits_{x 	o 0} {{x\cot \left( {4x} ight)} \over {{{\sin }^2}x{{\cot }^2}\left( {2x} ight)}} is equal to :

Accepted Answer

Answer: C. 1 \mathop {\lim }\limits_{x 	o 0} {{x{{	an }^2}2x} \over {	an 4x{{\sin }^2}x}} = \mathop {\lim }\limits_{x 	o 0} {{x\left( {{{{{	an }^2}2x} \over {4{x^2}}}} ight)4{x^2}} \over {\left( {{{	an 4x} \over {4x}}} ight)4x\left( {{{{{\sin }^2}x} \over {{x^2}}}} ight){x^2}}} = 1

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

Answer

Explanation

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