lim n → ∞ sin π n 2 + 1 is equal to

lim n → ∞ sin n π 1 + 1 n 2 1 / 2 = lim n → ∞ sin n π 1 + 1 2 n 2 - 1 8 n 4 + … = lim n → ∞ sin n π + π 2 n - π 8 n 3 + … = lim n → ∞ - 1 n sin π 1 2 n - 1 8 n 3 + … = 0

Mathematics - Limits Question with Solution | TestHub

MathematicsLimitsGeneral ( Limit existence)Hard2 minPYQ_2017

MathematicsHardsingle choice

$\lim_{n \to \infty} \sin [π \sqrt{n^{2} + 1}]$ is equal to

Options:

Answer:

Solution:

$\lim_{n \to \infty} \sin \{n π {(1 + \frac{1}{n^{2}})}^{1 / 2}\}$

$= \lim_{n \to \infty} \sin \{n π (1 + \frac{1}{2 n^{2}} - \frac{1}{8 n^{4}} + \dots)\}$

$= \lim_{n \to \infty} \sin \{n π + \frac{π}{2 n} - \frac{π}{8 n^{3}} + \dots\}$

$= \lim_{n \to \infty} {(- 1)}^{n} \sin π (\frac{1}{2 n} - \frac{1}{8 n^{3}} + \dots) = 0$

Stream:BITSATSubject:MathematicsTopic:LimitsSubtopic:General ( Limit existence)

⏱ 2mℹ️ Source: PYQ_2017

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