Calculting limit $\cos(\sqrt{x+1})-\cos(\sqrt{x})$ as $x\to \infty$ I am not sure how to calculate the limit:
$$\lim_{x\to\infty}\cos(\sqrt{x+1})-\cos(\sqrt{x})$$
I applied trigonometric identity to get:
$$L=-\lim_{x\to \infty} 2\sin\left(\dfrac{1}{2}\dfrac{1}{(\sqrt{1+x}-\sqrt{x})}\right) 
\sin\left(\dfrac{1}{2}\dfrac{1}{(\sqrt{1+x}+\sqrt{x})}\right)$$
Not sure how to proceed using trigonometry.
Also, is there a method using L'Hospital for $\infty-\infty$ form (not for this question obviously!)?
 A: 
METHODOLOGY $1$: Pre-Calculus Approach 

Recalling that $\cos(x)-\cos(y)=-2\sin\left(\frac{x-y}{2}\right)\sin\left(\frac{x+y}{2}\right)$, $|\sin(\theta)|\le |\theta|$, and $|\sin(\theta)|\le 1$, we have
$$\begin{align}
\left|\cos(\sqrt{x+1})-\cos(\sqrt{x})\right|&=2\left|\sin\left(\frac{\sqrt{x+1}-\sqrt{x}}{2}\right)\sin\left(\frac{\sqrt{x+1}+\sqrt{x}}{2}\right)\right|\\\\
&=2\left|\underbrace{\sin\left(\frac{1}{2\left(\sqrt{x+1}+\sqrt{x}\right)}\right)}_{\text{bounded by its argument in absolute value}}\,\,\,\underbrace{\sin\left(\frac{\sqrt{x+1}+\sqrt{x}}{2}\right)}_{\text{bounded by}\,1 \,\text{in absolute value}}\right|\\\\
&\le \frac{1}{\sqrt{x+1}+\sqrt{x}}\\\\
&\le \frac{1}{2\sqrt{x}}\to 0
\end{align}$$
as $x\to \infty$.


METHODOLOGY $2$: Calculus Approach 

Alternatively, we can use the mean value theorem applied to $\cos(\sqrt{x})$.  Then, there exists a number $\xi\in (x,x+1)$ such that 
$$\cos(\sqrt{x+1})-\cos(\sqrt{x})=-\frac{\sin(\sqrt{\xi})}{2\sqrt{\xi}}\to 0$$
as $x\to \infty$.
