Calculate $\lim\limits_{n \to \infty} \cos(\sqrt{E((2n\pi)^2)\ }$ where $E(x)=\lfloor x\rfloor$ and $\lfloor .\rfloor$ refers to the floor function (and n is an integer).
I've tried to use that if such an expression has a limit, then it will equal its lim sup, and then use the inequality: $$4\pi^2n^2-1\leq \lfloor 4\pi^2n^2\rfloor \leq 4\pi^2n^2 \ \ \forall n \in \mathbb{N}^*$$ But the problem is there is no reason $\cos$ would be monotonic so that I could compose this inequality (and I'm not sure it would really permit me to finish a proof).
How can I continue?
We can rewrite the limit as
$$\lim_{n\to\infty}\cos\sqrt{4\pi^2n^2-\epsilon(n)}$$
where $0 < \epsilon < 1$ for some function $\epsilon(n)$ ($\epsilon(n)$ is the fractional part function, $\{\cdot\}$, but only this property is relevant - the inequalities are strict because $4\pi^2n^2$ is never an integer for $n>0$). Then for all $n$ we have that
$$\cos\sqrt{4\pi^2n^2-\epsilon(n)} = \cos\left(2\pi n \sqrt{1-\frac{\epsilon(n)}{4\pi^2n^2}}\right) = \cos\left(2\pi n - \frac{1}{4\pi n} + O(n^{-2})\right)$$ $$ = \cos\left(\frac{1}{4\pi n} + O(n^{-2})\right)$$
from the Taylor series for $\sqrt{1-x}$ and the cosine angle addition formula. Thus taking the limit we get $\cos(0) = 1$