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JEE Main 20154 April
Exam
JEE Main
Year
2015
Date
4 April
Subject
Mathematics
Chapter
Limits

JEE Main 2015 · 4 April · Mathematics · Limits

_ x → 0 ((1 - 2x)(3 + cos x))/(x 4x) =

lim⁡x→0(1−cos⁡2x)(3+cosx)xtan⁡4x=\lim_{x \to 0}\dfrac{\left(1 - \cos 2x\right)\left(3 + cos x\right)}{x \tan 4x} =

Options

  1. A12\dfrac{1}{2}
  2. B44
  3. C33
  4. D22

Correct answer

D. 22

Step-by-step solution

lim⁡x→0(1−cos⁡2x)(3+cosx)xtan4x\lim_{x \to 0} \dfrac{( 1 - \cos 2x ) ( 3 + cos x )}{xtan 4x}

=lim⁡x→02sin⁡2x⋅(3+cosx)x⋅tan⁡4x= \lim_{x \to 0} \dfrac{2\sin^{2} x ⋅ ( 3 + cos x )}{x ⋅ \tan 4x}

=lim⁡x→02⋅(sinxx)2⋅(3+cos⁡x)tan⁡4x4x⋅4= \lim_{x \to 0} \dfrac{2 ⋅ {\left(\dfrac{sin x}{x}\right)}^{2} ⋅ \left(3 + \cos x\right)}{\dfrac{\tan 4x}{4x} ⋅ 4}

=2⋅(1)2⋅(3+1)1⋅4= \dfrac{2 ⋅ {\left(1\right)}^{2} ⋅ ( 3 + 1 )}{1 ⋅ 4}

=2= 2 .
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