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JEE Main 20146 April
Exam
JEE Main
Year
2014
Date
6 April
Subject
Mathematics
Chapter
Limits

JEE Main 2014 · 6 April · Mathematics · Limits

_ x → 0 sin ( π cos ^ 2 x) x ^ 2 is equal to

\lim_{x \to 0}\dfrac{sin\left({\pi cos}^{2}x\right)}{{x}^{2}} is equal to

Options

  1. A- π
  2. Bπ\pi
  3. C\dfrac{π}{2}
  4. D1

Correct answer

B. π\pi

Step-by-step solution

We have lim⁡x→0sin⁡(πcos2x)x2\lim_{x \to 0}\dfrac{\sin \left({\pi cos}^{2}x\right)}{{x}^{2}}

= \lim_{x \to 0}\dfrac{\sin \left(\pi - {\pi cos}^{2}x\right)}{{x}^{2}} \left[∵ \text\sin \left(\pi - \theta\right) = \text\sin \theta\right]

=lim⁡x→0sin⁡(πsin⁡2x)(πsin⁡2x)×πsin⁡2xx2\text{=}\lim_{x \to 0}\dfrac{\sin \left(\pi\sin^{2}x\right)}{\left(\pi\sin^{2}x\right)} \times \dfrac{\pi\sin^{2}x}{{x}^{2}}

\text{=}\lim_{x \to 0}\pi\left[\dfrac{\sin \left(\pi\sin^{2}x\right)}{\left(\pi\sin^{2}x\right)}\right] {\left[\dfrac{\text\sin \text{x}}{\text{x}}\right]}^{2}

=π(1)(1)[As,lim⁡x→0sin⁡tt=1]= \pi\left(1\right)\left(1\right) \left[As, \lim_{x \to 0}\dfrac{\sin t}{t} = 1\right]

=π= \pi.
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