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Determine the following limit, or show it doesn't exist: $$\lim_{n\to \infty}(\sin\sqrt{n+1} - \sin\sqrt{n}) .$$

I'm not sure how to proceed. I know that I can't use limit arithmetic because both $\lim_{n\to \infty}\sin\sqrt{n+1}$ and $\lim_{n\to \infty}\sin\sqrt{n}$ diverge, although I'm not really sure that fact is all that useful in solving this.

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    $\begingroup$ use the mean value theorem $\endgroup$ – user124910 Dec 18 '18 at 4:05
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    $\begingroup$ $\sin\alpha-\sin\beta=\ldots$ $\endgroup$ – Artem Dec 18 '18 at 4:10
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Hint Since $\sin$ is differentiable and $| {\sin x} | \leq 1$ for all (real) $x$, we have $$|\sin x - \sin y| \leq |x - y| .$$

Taking $x = \sqrt{n + 1}$ and $y = \sqrt{n}$ gives gives that the quantity $\sin \sqrt{n + 1} - \sin \sqrt{n}$ whose limit we're evaluating is bounded above in absolute value by $$\sqrt{n + 1} - \sqrt{n} = \frac{1}{\sqrt{n + 1} + \sqrt{n}} \leq \frac{1}{2 \sqrt{n}} .$$

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$\sin x < x $
$|\sin \sqrt{x+1} - sin \sqrt{x}| \leq |\sqrt{x+1} - \sqrt{x}| = \dfrac{1}{|\sqrt{x+1} + \sqrt{x}|} \to 0$

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$$\sin\sqrt{x+1}-\sin\sqrt x=2\sin\dfrac{\sqrt{x+1}-\sqrt x}2\cos\dfrac{\sqrt{x+1}+\sqrt x}2$$

For real $y,-1\le\cos y\le1$

For $\lim_{x\to\infty}\sin(...)=\lim_{...}\sin\dfrac1{2(\sqrt{x+1}+\sqrt x)}=?$

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By the mean value theorem,

$|\sin(\sqrt{n+1}) - \sin(\sqrt{n})| \leq |\cos(\theta_n)||\sqrt{n+1}-\sqrt{n}|$ and $|\cos(\theta_n)| \leq 1$ for all n. Now its easy to finish.

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