Limit inferior of functions at infinity How do you define 
$$\liminf_{x\to\infty} f(x) <a $$
where $a>0$, and $f$ is continuous on $\mathbb R$?
Is it the following?  

There exist $x_{0}$ such that $f(x)<a$ for all $x>x_{0}$

 A: By definition, there must exist a sequence $\{x_n\}_n$ such that $x_n \to +\infty$ and $$\lim_{n \to +\infty} f(x_n) < a.$$
Indeed, $$\liminf_{x \to +\infty} f(x) = \sup_{N \in \mathcal{U}(+\infty)} \inf_{x \in N} f(x),$$ where $\mathcal{U}(+\infty)$ is the system of neighborhoods of $+\infty$. You conclude by considering a countable family of neighborhoods like $\{(n,+\infty)\}_{n \in \mathbb{N}}$.
A: No. By definition $$\liminf_{x\to\infty}f(x)=\lim_{x_0\to\infty}\inf_{x\ge x_0}f(x)\;.$$
Consider the function 
$$f(x)=\begin{cases}
1,&\text{if }x\le 1\\\\
\frac1n-(x-n)\left(\frac1n+\frac1{n+1}\right),&\text{if }n\in\Bbb Z^+\text{ is odd and }n\le x<n+1\\\\
\frac{-1}n+(x-n)\left(\frac1n+\frac1{n+1}\right),&\text{if }n\in\Bbb Z^+\text{ is even and }n\le x<n+1\;;
\end{cases}$$
$f$ is piecewise linear, and for $x\ge 1$ its graph is a sawtooth with upper points at $\left\langle n,\frac1n\right\rangle$ for odd $n$ and lower points at $\left\langle n,\frac{-1}n\right\rangle$ for even $n$.
Fix some positive $x_0\in\Bbb R$. Let $2n$ be the smallest even integer greater than or equal to $x_0$; then $$\inf_{x\ge x_0}f(x)=\frac{-1}{2n}\;,$$ because all of the troughs to the right of this one are shallower. (That is, $f(x)$ has local minima at $2n$ for $n\in\Bbb Z^+$, and these minima get larger as $n$ increases.) But then $$\liminf_{x\to\infty}f(x)=\lim_{n\to\infty}\frac{-1}{2n}=0\;.$$ On the other hand, for every $x_0\in\Bbb R$ there are $x>x_0$ such that $f(x)<0$, so it’s not true that $f(x)\ge 0$ for all sufficiently large $x$.
A: 
There exists $\varepsilon\gt0$ such that $\{x\mid f(x)\leqslant a-\varepsilon\}$ is unbounded. 

Somewhat more explicitly: There exists $\varepsilon\gt0$ such that, for every finite $x_0$, there exists $x\geqslant x_0$ such that $f(x)\leqslant a-\varepsilon$.
Equivalently: There exists $\varepsilon\gt0$ such that $\{x\mid f(x)\lt a-\varepsilon\}$ is unbounded.
