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

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.

• use the mean value theorem – user124910 Dec 18 '18 at 4:05
• $\sin\alpha-\sin\beta=\ldots$ – Artem Dec 18 '18 at 4:10

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 $$\sqrt{n + 1} - \sqrt{n} = \frac{1}{\sqrt{n + 1} + \sqrt{n}} \leq \frac{1}{2 \sqrt{n}} .$$

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

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

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

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

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.