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JEE Main 201911 JanuaryEvening Shift
Exam
JEE Main
Year
2019
Date
11 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Limits

JEE Main 2019 · 11 January · Evening Shift · Mathematics · Limits

_ x arrow 0 x (4 x) ^ 2 x ^ 2 (2 x) is equal to:

lim⁡x→0xcot⁡(4x)sin⁡2xcot⁡2(2x)\lim _{x \rightarrow 0} \frac{x \cot (4 x)}{\sin ^{2} x \cot ^{2}(2 x)} is equal to:

Options

  1. A0
  2. B2
  3. C4
  4. D1

Correct answer

D. 1

Step-by-step solution

lim⁡x→0xcot⁡4xsin⁡2x⋅cot⁡22x=lim⁡x→0x⋅tan⁡22xsin⁡2x⋅tan⁡4x\lim _{x \rightarrow 0} \frac{x \cot 4 x}{\sin ^{2} x \cdot \cot ^{2} 2 x}=\lim _{x \rightarrow 0} \frac{x \cdot \tan ^{2} 2 x}{\sin ^{2} x \cdot \tan 4 x} =lim⁡x→0(xsin⁡x)2⋅(tan⁡2x2x)2⋅(4xtan⁡4x)⋅422=1=\lim _{x \rightarrow 0}\left(\frac{x}{\sin x}\right)^{2} \cdot\left(\frac{\tan 2 x}{2 x}\right)^{2} \cdot\left(\frac{4 x}{\tan 4 x}\right) \cdot \frac{4}{2^{2}}=1
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