Limit $\lim_{x \to \infty}{\sin{\sqrt{x+1}}-\sin{\sqrt{x}}}$ I want to compute $$\lim_{x \to \infty}{\sin{\sqrt{x+1}}-\sin{\sqrt{x}}}.$$
Is it OK how I want to do?  
$$\sin{\sqrt{x+1}}-\sin{\sqrt{x}}=2\sin{\frac{\sqrt{x+1}-\sqrt{x}}{2}}\cos{\frac{\sqrt{x+1}+\sqrt{x}}{2}}=2\sin{\frac{1}{2(\sqrt{x+1}+\sqrt{x})}}\cos{\frac{\sqrt{x+1}+\sqrt{x}}{2}}$$
I'm not surem but I want to say that $$|\cos \frac{\sqrt{x+1}-\sqrt{x}}{2}| \leq 1$$ $$\sin{\frac{1}{2(\sqrt{x+1}+\sqrt{x})}} \to 0 \mbox{ when } x \to \infty$$
So the limit is $0$ ?
I thinks is not because  if I say that $ \displaystyle|\sin{\frac{1}{2(\sqrt{x+1}+\sqrt{x})}}| \leq 1$ I will obtain that another limit than $0$. 
Thanks :) 

Assuming you have the Mean Value Theorem at your dispoeal, that's the easiest way to show that the limit equals $0$: with $g(x) = \sin\sqrt x$, we have
$$
\sin\sqrt{x+1}-\sin\sqrt x = g(x+1)-g(x) = ((x+1)-x)g'(\xi) = \frac{\cos\sqrt\xi}{2\sqrt\xi}
$$
for some $x\le\xi\le x+1$. In particular,
$$
|\sin\sqrt{x+1}-\sin\sqrt x| = \frac{|\cos\sqrt\xi|}{|2\sqrt\xi|} \le \frac1{2\sqrt x},
$$
and so $\lim_{x\to\infty} (\sin\sqrt{x+1}-\sin\sqrt x) = 0$ by the Squeeze Theorem.
