Evaluate $\lim\limits_{x\to\infty}(\sin\sqrt{x+1}-\sin\sqrt{x})$ When using Maclaurin series, the limit is
$$\lim\limits_{x\to\infty}\frac{1}{\sqrt{x+1}+\sqrt{x}}=0$$
If we expand the expression with two limits
$$\lim\limits_{x\to\infty}\sin\sqrt{x+1}-\lim\limits_{x\to\infty}\sin\sqrt{x}$$
it diverges.
Which solution is right?

Hint:
Mean value theorem implies there exist $c\in]x,x+1[$ such that
$$\sin(\sqrt{x+1})-\sin(\sqrt{x})=\frac{\sqrt{x+1}-\sqrt{x}}{2\sqrt{c}}\cos\sqrt{c}$$
Then $$\left|\sin(\sqrt{x+1})-\sin(\sqrt{x})\right|\le\left(\frac{\sqrt{x+1}}{2\sqrt{x}}-\frac{\sqrt{x}}{2\sqrt{x+1}}\right)|\cos\sqrt{c}|\le\frac{1}{2}\left(\sqrt{1+\frac{1}{x}}-\sqrt{\frac{1}{1+1/x}}\right)$$
