TITLE: Proving $\lim_{x\to 0}f(x)=\infty$.
QUESTION [5 upvotes]: Suppose we have a function $f:[0,\infty)\to \mathbb{R}$ such that for every $N\in\Bbb{N}$ and every sequence of $\delta_n>0$ such that $\lim_{n\to\infty}\delta_n=0$, there exists $n$ for which $f(\delta_n)\geq N$.
Does that imply that
$$\lim_{x\to 0}f(x)=\infty?$$
I am confused about this, because I would first think we must have $f(\delta_n)\geq N$ for all $n$ large enough, to conclude that $f(x)\to\infty$. But here for each sequence $\delta_n$ we only have one $n$ for which $f(\delta_n)\geq N$. But I cannot think of any counterexample, so maybe the statement is true. How to prove it?

REPLY [3 votes]: The fact that the statement works over all sequences $\{\delta_n\}$ is very strong.  Note that for any suitable sequence $\{\delta_n\}$, consider the subset of those $n\in \Bbb N$ such that $f(\delta_n) < N$, then that set is finite.  Otherwise, let the set be of the form $\{n_1, n_2, \ldots\}$ and now consider instead the sequence $\{\delta_{n_i}\}$.  Note that $\delta_{n_i} \to 0$ so we get some $k$ such that $(\delta_{n_k}) \ge N$, a contradiction.
The desired limit is indeed $\infty$, and to see this, assume the negation holds.  This means that there is some $N$ such that for every $\delta>0$ there is some $x$ such that $x\in [0, \delta)$ and $f(x) < N$.  It should be clear how to generate a contradictory sequence $\{\delta_n\}$ from this information.