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$\lim (-1)^n \sin\bigl(\pi(n^2+0.5n+1)^{\frac{1}{2}}\bigr)\sin\Bigl(\frac{(n+1)\pi}{4n}\Bigr)$

Where tends to infinity

The function seems to complex so any hint to start with it

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  • $\begingroup$ When n -> infty, n^2+0.5n+1 is approx n^2, so the argument to the first sine is n pi. The second one has (n+1)/n which tends to 1. Of course, a formal demonstration requires a bit more care. $\endgroup$ Mar 29, 2021 at 13:23

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$$\sin(\pi\sqrt{n^2 +0.5n +1} )=\sin\left(\pi n\sqrt{1+\frac{1}{2n} +\frac{1}{n^2}}\right ) $$ Now expand the square root binomially, and ignore the higher order terms as they all $\to 0$. $$\to \sin\left(\pi n\left(1+\frac 12\left(\frac{1}{2n} +\frac{1}{n^2} \right) \right) \right) \\ \to \sin(n\pi +\frac{\pi}{4} ) =(-1)^n \frac{1}{\sqrt 2}$$ In the second part of the limit, just notice that $\frac{n+1}{n} \to 1 $.

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