Two years ago I found this question and wasn't able to make the slightest advancement.
$\lim_{n\to \infty} \cos(n)^{(n²)} $
The question is does the limit exist, where n is natural only (of course it does not exist for n being real).
Up until now, I noticed the limit does not exist only if there are peaks of cosine in natural numbers, that peaks "more" than raising to n² power.
Since cosine peaks in $2\pi k$, the question arising is "how good can we approximate multiples of $\pi$ using a natural number, or asking what is the best rational approximation for $\pi$.
In fact, this limit does not exist. Let $a_n=\cos^{ \ n^2}{n}$, by using Hurwitz's theorem, there are infinitely many relatively prime integers $p,q$ such that \begin{align*} \left|\pi-\frac{p}{q}\right|<\frac{1}{\sqrt{5}q^2}<\frac{1}{q^2}\Rightarrow \left|p-q\pi\right|<\frac{1}{q} \end{align*} arrange the $\frac{p}{q}$ satisfying the above conditions into the sequence $\left\{\frac{p_n}{q_n}\right\}_{n=1}^\infty$ in the monotonically increasing order of $q_n$. In addition, there exist $N_1>0$ such that $\forall \ n>N_1, \ 3<\frac{p_n}{q_n}<4$ and \begin{align*} \left|\cos{p_n}-\cos\left(q_n\pi\right)\right|&<\left|\cos\left(q_n\pi+\frac{1}{q_n}\right)-\cos\left(q_n\pi\right)\right|\\ &=2\left|\sin\left(q_n\pi+\frac{1}{2q_n}\right)\sin{\frac{1}{2q_n}}\right|\\ &=2\sin^2\frac{1}{2q_n}<\frac{1}{2q_n^2}<\frac{8}{p^2_n} \end{align*} by using triangle inequality we have \begin{align*} 1-\left|\cos{p_n}\right|=\left|\left|\cos{p_n}\right|-\left|\cos\left(q_n\pi\right)\right|\right|\le \left|\cos{p_n}-\cos\left(q_n\pi\right)\right|<\frac{8}{p_n^2} \end{align*} so $\left|\cos{p_n}\right|>1-8/p_n^2$. When $n$ is large enough to make $1-8/p_n^2>0$, we get \begin{align*} \left|a_{p_n}\right|=\left| \cos^{ \ p_n^2}{p_n}\right|>\left(1-\frac{8}{p_n^2}\right)^{p_n^2}\to e^{-8},\quad n\to \infty \end{align*} this means that there exist $N_2>0$ such that \begin{align*} \left|a_{p_n}\right|>10^{-8},\quad \forall \ n>N_2\tag{1} \end{align*} since $10^{-8}<e^{-8}$. On the other hand, Weyl equidistribution theorem implies that $\left\{n\pi\operatorname*{mod} 1: n\in \mathbb{N}^*\right\}$ is dense on the interval $(0,1)$, so there are infinitely many $n$ and a fixed $\epsilon\in (0,1/2)$ such that \begin{align*} \epsilon< n\pi \operatorname*{mod} 1<1-\epsilon \end{align*} arrange the $n$ satisfying the above conditions into the sequence $\left\{n_k\right\}_{k=1}^\infty$ in the monotonically increasing order of $n_k$. Let $h_k=\left[n_k\pi\right]$, so $\left|\cos{h_k}\right|<\cos{\epsilon}$ and \begin{align*} \lim_{k\to\infty}\left|\cos^{ \ h_k^2}{h_k}\right|\le \lim_{k\to\infty}\left|\cos{\epsilon}\right|^{ \ h_k^2}=0\Rightarrow \lim_{k\to\infty}a_{h_k}=0\tag{2}. \end{align*} Combining $(1)$ with $(2)$, we have that the limit of the sequence $a_n$ does not exist.