Functions or sequence divergences link

Question

One can prove that the sequence $ u_n=\{\sin(n)\}_{n \in \mathbb{N}} $ diverges using a similar argument as in : Proves the divergence of sequence of sin(n)
But we can also prove that $f(x)=\sin(x)$ diverges at $x=+\infty$ because it's continuous and has two different limit 1 and -1 at $ x_n=\frac{(4n+1)\pi}{2} $ and $y_n=\frac{(4n+3)\pi}{2}$

Now my question is :
Can we use the latter fact to conclude that the sequence diverges ? In fact $x_n$ and $y_n$ are not integers, so $\sin(x_n)$ or $\sin(y_n)$ are not subsequences of the initial sequence, so that's maybe not enough to conclude it for that only reason.
In other word, what conditions are required (continuity, etc), to conclude that if $f(x)$ diverges at $x=+\infty$ then $u_n = f(n)$ diverges as well.

Thanks !

Answer 1

None that I can think of. Let $f\colon[a,\infty)\to\mathbb{R}$ be continuous and such that $$ l=\liminf_{x\to\infty}f(x)<\limsup_{x\to\infty}f(x)=L. $$ Then for any $y\in(l,L)$ there is a sequence $\{x_n\}$ such that $$ \lim_{n\to\infty}x_n=\infty\quad\text{and}\quad f(x_n)=y\quad\forall n\in\mathbb{N}. $$ For instance, in your example, $f(x)=\sin x$, $l=-1$ and $L=1$. If $y=1/2$, we can take $x_n=\pi/6+2\,n\,\pi$.

The moral of the story is that from the divergence of $f(x)$ as $x\to\infty$ you cannot deduce nothing about the convergence or not of a particular sequence $\{f(x_n)\}$, $x_n\to\infty$. On the other hand, if $f(x)$ converges as $x\to\infty$, then any sequence $\{f(x_n)\}$ with $x_n\to\infty$ converges to the same limit.