Proving $\lim_{x\to 0}f(x)=\infty$.

Question

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?

If the limit exists it certainly must be $\infty$. I'm still trying to decide whether or not the limit must exist at all. — TravisJ, Apr 18 '15 at 01:31

score 3 · Accepted Answer · answered Apr 18 '15 at 01:39

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.

Proving $\lim_{x\to 0}f(x)=\infty$.

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