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JEE Main 201510 AprilOnline
Exam
JEE Main
Year
2015
Date
10 April
Mode
Online
Subject
Mathematics
Chapter
Limits

JEE Main 2015 · 10 April · Online · Mathematics · Limits

_ x → 0 e ^ x ^ 2 - cos x ^ 2 x is equal to

lim⁡x→0ex2−cosxsin⁡2x\lim_{x \to 0} \dfrac{{e}^{{x}^{2}} - cos x}{\sin^{2} x}is equal to

Options

  1. A22
  2. B32\dfrac{3}{2}
  3. C54\dfrac{5}{4}
  4. D33

Correct answer

B. 32\dfrac{3}{2}

Step-by-step solution

lim⁡x→0ex2−cos⁡xsin⁡2x=lim⁡x→0(ex2−1)+(1−cos⁡x)x2⋅(sin⁡2xx2)\lim_{x \to 0}\dfrac{{e}^{{x}^{2}} - \cos x}{\sin^{2}x} = \lim_{x \to 0}\dfrac{( {e}^{{x}^{2}} - 1 ) + ( 1 - \cos x )}{{x}^{2} \cdot \left(\dfrac{\sin^{2}x}{{x}^{2}}\right)}

=lim⁡x→0(ex2−1x2+1−cos⁡xx2)= \lim_{x \to 0}\left(\dfrac{{e}^{{x}^{2}} - 1}{{x}^{2}} + \dfrac{1 - \cos x}{{x}^{2}}\right)

=1+12=32= 1 + \dfrac{1}{2} = \dfrac{3}{2}
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