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Algebra of Limits

Evaluate n ...

Question

Evaluate $lim n \to \infty n sin (2 π \sqrt{1 + n^{2}}), (n \in N)$

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Solution

$l i m n \to \infty n sin (2 π \sqrt{n^{2} + 1})$

$= l i m n \to \infty n sin (2 n π \sqrt{1 + \frac{1}{n^{2}}})$

$= l i m n \to \infty n sin (2 n π \sqrt{1 + 0})$

$= l i m n \to \infty n sin (2 n π)$

$= l i m n \to \infty n \times 0$

$= l i m n \to \infty 0$

$= 0$

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