Solving $\lim_{x\to+\infty}(\sin\sqrt{x+1}-\sin\sqrt{x})$ Do you have any tips on how to solve the limit in the title? Whatever I think of doesn't lead to the solution. I tried using: $\sin{x}-\sin{y}=2\cos{\frac{x+y}{2}}\sin{\frac{x-y}{2}}$ and I got:
$$\lim_{x\to+\infty}\bigg(2\cos{\frac{\sqrt{x+1}+\sqrt{x}}{2}}\sin{\frac{\sqrt{x+1}-\sqrt{x}}{2}}\bigg)=$$
$$\lim_{x\to+\infty}\bigg(2\cos{\frac{\sqrt{x+1}+\sqrt{x}}{2}\frac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x+1}-\sqrt{x}}}\sin{\frac{\sqrt{x+1}-\sqrt{x}}{2}}\bigg)=$$
$$\lim_{x\to+\infty}\bigg(2\cos{\frac{1}{2(\sqrt{x+1}-\sqrt{x})}}\sin{\frac{\sqrt{x+1}-\sqrt{x}}{2}}\bigg)$$
but, as you can see, this leads me to $\infty-\infty$ in $\cos$ term. How can I get rid of that?
 A: Your difference of sines method is good. Leave the cosine term alone, and work with the sine term. 
Note that $\frac{\sqrt{x+1}-\sqrt{x}}{2}=\frac{1}{2(\sqrt{x+1}+\sqrt{x})}$. This goes t0 $0$ as $x\to\infty$, so the sine term goes to $0$. More informally, $\sqrt{x+1}-\sqrt{x}$ is close to $0$ when $x$ is large.
The cosine term stays between $-1$ and $1$, so the product has limit $0$.
A: I thought it might be instructive to present a very efficient way forward.  From the Mean Value Theorem, with $f(x)=\sin\left(\sqrt{x}\right)$ and $f'(x)=\frac{\cos\left(\sqrt{x}\right)}{2\sqrt{x}}$, there exists a number $x< \xi< x+1$ such that 
$$\sin\left(\sqrt{x+1}\right)-\sin\left(\sqrt{x}\right)=\frac{\cos\left(\sqrt{\xi}\right)}{2\sqrt{\xi}}$$
Then, taking a limit as $x\to \infty$, $\xi \to \infty$ since $\xi\in (x,x+1)$, and we see immediately that 
$$\begin{align}
\lim_{x\to \infty}\left(\sin\left(\sqrt{x+1}\right)-\sin\left(\sqrt{x}\right)\right)&=\lim_{\xi \to \infty}\frac{\cos\left(\sqrt{\xi}\right)}{2\sqrt{\xi}}\\\\
&=0
\end{align}$$
A: Notice, $$\lim_{x\to +\infty}\left(2\cos\left(\frac{1}{2(\sqrt{x+1}-\sqrt x)}\right)\sin\left(\frac{ \sqrt{x+1}-\sqrt x}{2}\right) \right)$$
$$=2\lim_{x\to +\infty}\cos\left(\frac{1}{2(\sqrt{x+1}-\sqrt x)}\right)\sin\left(\frac{1}{2(\sqrt{x+1}+\sqrt x)}\right)$$
$$=2\lim_{x\to +\infty}\cos\left(\frac{1}{2(\sqrt{x+1}-\sqrt x)}\right)\cdot \lim_{x\to +\infty}\sin\left(\frac{1}{2(\sqrt{x+1}+\sqrt x)}\right)$$
since, $-1\le \cos y\le 1\ \ \ \forall \ \ \ y\in R$, 
$$=2(k)\cdot \sin\left(0\right)=\color{red}{0}$$
where, $-1\le k\le 1$
