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BITSAT 2017
Exam
BITSAT
Year
2017
Subject
Mathematics
Chapter
Limits

BITSAT 2017 · Mathematics · Limits

_ n → ∞ [π n^ 2 + 1 ] is equal to

lim⁡n→∞sin⁡[πn2+1]isequalto\lim_{n \to \infty}\sin \left[\pi\sqrt{n^{2} + 1}\right] is equal to

Options

  1. A∞\infty
  2. B00
  3. Cdoes not exist
  4. DNone of these

Correct answer

B. 00

Step-by-step solution

lim⁡n→∞sin⁡{nπ(1+1n2)1/2}\lim_{n \to \infty}\sin \left\{n\pi\left(1 + \dfrac{1}{n2}\right)^{1} / 2\right\}

=lim⁡n→∞sin⁡{nπ(1+12n2−18n4+…)}= \lim_{n \to \infty}\sin \left\{n\pi\left(1 + \dfrac{1}{2n^{2}} - \dfrac{1}{8n^{4}} + \ldots\right)\right\}

=lim⁡n→∞sin⁡{nπ+π2n−π8n3+…}= \lim_{n \to \infty}\sin \left\{n\pi + \dfrac{\pi}{2n} - \dfrac{\pi}{8n^{3}} + \ldots\right\}

=lim⁡n→∞(−1)nsinπ(12n−18n3+…)=0= \lim_{n \to \infty}\left(- 1\right)nsin\pi\left(\dfrac{1}{2n} - \dfrac{1}{8n^{3}} + \ldots\right) = 0

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