Limn-sin -n2-n-1 - where n -N-

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Standard XII

Mathematics

Substitution Method to Remove Indeterminate Form

limn→∞|sin π√...

Question

$lim n \to \infty ∣ ∣ sin (π \sqrt{n^{2} + n + 1}) ∣ ∣ = ?$ where n $\in$ N.

1

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0

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Indeterminate

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Finite

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Solution

The correct options are
A $1$
D Finite
We know that, $| sin (θ) | = | sin (θ - n π) |$
$| sin (π \sqrt{n^{2} + n + 1}) | = | sin (π \sqrt{n^{2} + n + 1} - n π) |$
$∣ ∣ ∣ ∣ sin (π (\frac{\sqrt{n^{2} + n + 1} - n}{1})) ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ ∣ sin ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{π ({(\sqrt{n^{2} + n + 1})}^{2} - n^{2})}{(\sqrt{n^{2} + n + 1} + n)} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ ∣ ∣ ∣ ∣ ∣ ∣$
$∣ ∣ ∣ ∣ sin (π (\frac{n + 1}{\sqrt{n^{2} + n + 1} + n})) ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ ∣ sin ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{π (1 + \frac{1}{n})}{\sqrt{1 + \frac{1}{n} + \frac{1}{n^{2}}} + 1} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ ∣ ∣ ∣ ∣ ∣ ∣$
$\Rightarrow$ Required limit, $L = ∣ ∣ ∣ ∣ sin (\frac{π (1 + 0)}{\sqrt{1} + 1}) ∣ ∣ ∣ ∣ = ∣ ∣ ∣ sin \frac{π}{2} ∣ ∣ ∣ = 1$

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limn→∞∣∣sin(π√n2+n+1)∣∣=? where n∈N.

$lim n \to \infty ∣ ∣ sin (π \sqrt{n^{2} + n + 1}) ∣ ∣ = ?$ where n $\in$ N.