Evaluate $\lim_{x\to \infty} \cos (\sqrt {x+1})-\cos (\sqrt x)$ 
Evaluate $$\lim_{x\to \infty} \cos (\sqrt {x+1})-\cos (\sqrt x)$$

For this question I have tried using the squeeze theorem and some trigonometric manipulations such as using $$2\sin A\sin B=\cos (A-B) -\cos (A+B)$$. 
But they were of no use. I also tried using the substitutions like $$x=1/y$$ Hence as $x\to \infty$ , $y\to 0$ but still no success.
I could've also tried using L'Hospital rule but I am not able to convert the function in appropriate indeterminate form. Any help would be greatly appreciated 
 A: By your identity you have 
$$=2\sin\frac{\sqrt{x+1}+\sqrt{x}}{2}\sin\frac{\sqrt{x}-\sqrt{x+1}}{2}$$
Now$$\frac{\sqrt{x}-\sqrt{x+1}}{2}\to 0$$
And thus $$\sin\frac{\sqrt{x}-\sqrt{x+1}}{2}\to 0$$
However $2\sin\frac{\sqrt{x+1}+\sqrt{x}}{2}$ is bounded and so the limit is zero.
A: By the mean value theorem, for any $x$ there is a $c\in(\sqrt{x},\sqrt{x+1})$ such that:
$$\cos\sqrt{x+1}-\cos\sqrt{x} = -\sin(c)\left(\sqrt{x+1}-\sqrt{x}\right).$$
And so:
$$\left|\cos\sqrt{x+1}-\cos\sqrt{x}\right|\leq\sqrt{x+1}-\sqrt{x}=\frac{1}{\sqrt{x}+\sqrt{x+1}}\to 0$$

You don't need differentiability. 
Claim:
If  $f$ is uniformly continuous, and $\lim_{x\to\infty} (a(x)-b(x))=0$, you can show $f(a(x))-f(b(x))\to 0.$
Proof:
Given $\epsilon>0,$ we find $\delta$ such that if $|x-y|<\delta$ then $|f(x)-f(y)<\epsilon.$ (Definition of uniform continuity.)
Then we find $N$ such that when $x>N$, $|a(x)-b(x)|<\delta$.
So if $x>N$, $|a(x)-b(x)|<\delta$ so $|f(a(x))-f(b(x))|<\epsilon,$ and hence $$\lim_{x\to\infty}\left(f\left(a(x)\right)-f\left(b(x)\right)\right)=0.$$

In particular, any continuous and periodic function (like $\cos(x)$) on $\mathbb R$ is uniformly continuous.
And if $a(x)=\sqrt{x+1}$ and $b(x)=\sqrt{x},$ then $a(x)-b(x)=\frac{1}{\sqrt{x+1}+\sqrt{x}}\to 0.$
A: Note that
\begin{align*}
\lim_{x\to \infty} \sin\frac{\sqrt{x+1}-\sqrt{x}}{2}&=\lim_{x\to \infty} \sin\frac{1}{2(\sqrt{x+1}+\sqrt{x})}=0
\end{align*}
Since
\begin{align*}
\left|\cos (\sqrt {x+1})-\cos (\sqrt x)\right|&=\left|-2\sin\frac{\sqrt{x+1}+\sqrt{x}}{2}\sin\frac{\sqrt{x+1}-\sqrt{x}}{2}\right|\\
&\le 2\left|\sin\frac{\sqrt{x+1}-\sqrt{x}}{2}\right|\\
&\to 0
\end{align*}
as $x\to\infty$,
\begin{align*}
\lim_{x\to \infty} \left[\cos (\sqrt {x+1})-\cos (\sqrt x)\right]&=0
\end{align*}
A: $$\lim_{x\rightarrow +\infty}\cos(\sqrt{x+1})-\cos(\sqrt{x})$$$$=\lim_{x\rightarrow +\infty}2\sin\left(\frac{\sqrt{x+1}+\sqrt x}{2}\right)\sin\left(\frac{\sqrt{x}-\sqrt{x+1}}{2}\right)$$
$$=-\lim_{x\rightarrow +\infty}\underbrace{\sin\left(\frac{\sqrt{1+x}+\sqrt x}{2}\right)}_{\to [-1, 1]}\cdot \underbrace{\frac{\sin\left(\frac{\sqrt{x+1}-\sqrt{x}}{2}\right)}{\left(\frac{\sqrt{x+1}-\sqrt x}{2}\right)}}_{\to 1}\cdot \underbrace{(\sqrt{x+1}-\sqrt x)}_{\to 0}=\color{blue}{0}$$
