A) 0
B) 2
C) 4
D) 1
Correct Answer: D
Solution :
Here,\[\underset{x\to 0}{\mathop{\lim }}\,\frac{x\,\cot (4x)}{{{\sin }^{2}}x{{\cot }^{2}}(2x)}=\underset{x\to 0}{\mathop{\lim }}\,\frac{x\,{{\tan }^{2}}2x}{\tan 4x{{\sin }^{2}}x}\] \[\underset{x\to 0}{\mathop{\lim }}\,\frac{x\left( \frac{{{\tan }^{2}}2x}{4{{x}^{2}}} \right)4{{x}^{2}}}{\left( \frac{\tan 4x}{4x} \right)4x\left( \frac{{{\sin }^{2}}x}{{{x}^{2}}} \right){{x}^{2}}}=1\]
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