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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

  1. A \(0\)
  2. B \(2\)
  3. C \(4\)
  4. D \(1\)
Verified Solution

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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