Limit of a sequence with a difference of cosines Define the following sequence:
$$x_n = \sqrt[3]{n+1}\cos{\sqrt{n+1}} - \sqrt[3]{n}\cos{\sqrt{n}},\forall n\in \mathbb{N} $$
Does the limit 
$$\lim_{n \to \infty} x_n$$ 
exist?
I believe that it does not, mainly because (i think that) $\cos\sqrt{n+1}$ and $\cos{\sqrt{n}}$ have opposite signs an infinite number of times.
More exactly, I tried showing that there exists an infinite sequence $a_n$ of natural numbers such that 
$$ \sqrt{a_n} \text{ mod } 2\pi \in \left(\frac{3\pi}{2} + \frac{\pi}{4} ,2\pi\right) \text{ and } \sqrt{a_n + 1} \text{ mod } 2\pi \in \left( 0, \frac{\pi}{4} \right)$$
and a sequence $b_n$ such that
$$\sqrt{b_n} \text{ mod } 2\pi \in \left(\frac{\pi}{2} + \frac{\pi}{4}, \pi \right) \text{ and } \sqrt{b_n + 1} \text{ mod }\in \left(\pi, \pi + \frac{\pi}{4} \right),$$
because then the subsequences $x_{a_n}$ and $x_{b_n}$ would both be bounded and of different signs, showing that $x_n$ is divergent.
Would this approach work? Is the sequence actually somehow convergent?  

By binomial series
$$\sqrt[3]{n+1}=\sqrt[3]{n}\left(1+\frac1n\right)^\frac13=\sqrt[3]{n}+O\left(\frac1{n^\frac23}\right)$$
then
$$\sqrt[3]{n+1}\cos{\sqrt{n+1}} - \sqrt[3]{n}\cos{\sqrt{n}}=$$$$=\sqrt[3]{n}(\cos{\sqrt{n+1}}-\cos{\sqrt{n}})+O\left(\frac{\cos{\sqrt{n+1}}}{n^\frac23}\right)$$
and by sum to product formula
$$\cos{\sqrt{n+1}}-\cos{\sqrt{n}}=-2\sin\left(\frac{\sqrt{n+1}+\sqrt{n}}{2}\right)\sin\left(\frac{\sqrt{n+1}-\sqrt{n}}{2}\right)=$$
$$=-2\sin\left(\frac{\sqrt{n+1}+\sqrt{n}}{2}\right)\sin\left(\frac{1}{2(\sqrt{n+1}+\sqrt{n})}\right)\sim -\frac{\sin(\sqrt n)}{\sqrt n}$$
therefore the given sequence converges to zero.
