<s>[INST] 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 :) 
 [/INST] Here's another approach, using the OP's trick of getting the square roots into the denominator, but with a different trig identity, in this case $\sin(a+b)=\sin a\cos b+\cos a\sin b$.
$$\begin{align}
|\sin(\sqrt{x+1})-\sin\sqrt x|
&=|\sin(\sqrt x+\sqrt{x+1}-\sqrt x)-\sin\sqrt x|\\
&=|\sin\sqrt x ( \cos(\sqrt{x+1}-\sqrt x)-1)+\cos\sqrt x\sin(\sqrt{x-1}-\sqrt x)|\\
&\le|\cos(\sqrt{x+1}-\sqrt x)-1|+|\sin(\sqrt{x-1}-\sqrt x)|\\
&=\left|\cos\left(1\over\sqrt{x+1}+\sqrt x \right)-1\right|+\left|\sin\left(1\over\sqrt{x+1}+\sqrt x \right) \right|\\
&\to|\cos0-1|+|\sin0|=|1-1|+|0|=0
\end{align}$$
</s>