Q36 / 90 · JEE Main 2019 (11 Jan Shift 2) · Mathematics

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

  • A
    0
  • B
    2
  • C
    4
  • Correct answer
    D
    1

Solution

$\lim _{x \rightarrow 0} \frac{x \cot 4 x}{\sin ^{2} x \cdot \cot ^{2} 2 x}=\lim _{x \rightarrow 0} \frac{x \cdot \tan ^{2} 2 x}{\sin ^{2} x \cdot \tan 4 x}$ $=\lim _{x \rightarrow 0}\left(\frac{x}{\sin x}\right)^{2} \cdot\left(\frac{\tan 2 x}{2 x}\right)^{2} \cdot\left(\frac{4 x}{\tan 4 x}\right) \cdot \frac{4}{2^{2}}=1$

Hence, the correct answer is option: 'D'.

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