JEE Mains · Maths · STD 11 - 12. limits
\(\mathop {\lim }\limits_{x \to 0} \,\frac{{x\,\cot \,\left( {4x} \right)}}{{{{\sin }^2}\,x\,{{\cot }^2}\,\left( {2x} \right)}}\) is equal to
- A \(0\)
- B \(2\)
- C \(4\)
- D \(1\)
Answer & Solution
Correct Answer
(D) \(1\)
Step-by-step Solution
Detailed explanation
\(\frac{{x\cos 4x{{\sin }^2}2x}}{{{{\sin }^2}x.{{\cos }^2}2x.\sin 4x}}\) \( = \frac{{4x}}{{\sin 4x}}.\frac{{\cos 4x}}{{{{\cos }^2}2x}}{\cos ^2}x\) \( \Rightarrow 1\,\,\,\,\,as\,\,\,x \to 0\)
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