Does there exist a function $f: \Bbb{R} \to \Bbb{R}$ with $\lim_{x \to x_0}|f(x)|=\infty$ for every $x_0 \in \Bbb{R}$? A function  $f: \Bbb{R} \to \Bbb{R}$ is said to be anti-continuous at $x_0$ if $\lim_{x \to x_0}|f(x)|=\infty$. This is equivalent to taking the epsilon-delta definition for a function being continuous, but requiring that $|f(x) - f(x_0)| > \epsilon$ for $x$ sufficiently close to $x_0$.
Does there exist a function that is anti-continuous at all real values? Or even just a dense set of real values?
I have noticed that if we restrict the domain to $\Bbb{Q}$ then we can simply take the denominator of the fractions in simplest form.
 A: The answer is NO. We prove by contradiction. Suppose that there exists
a function $f:\mathbb{R}\rightarrow\mathbb{R}$ such that for each
$x_{0}\in\mathbb{R}$, $\lim_{x\rightarrow x_{0}}|f(x)|=\infty$.
For each $n\in\mathbb{N}$, define $A_{n}=\{x\in\mathbb{R}\mid|f(x)|\leq n\}$.
Note that $A_{n}$ does not have any accumulation point (If $A\subseteq\mathbb{R}$,
we say that $x\in\mathbb{R}$ is an accumulation point of $A$ if
there exists a sequence $(x_{n})$ in $A$ such that $x\neq x_{n}$
and $x_{n}\rightarrow x$.) because if $x_{0}$ is an accumulation
of $A_{n}$, we can choose a sequence $(x_{k})$ in $A_{n}$, with
$x_{k}\neq x_{0}$ and $x_{k}\rightarrow x_{0}$. Then $|f(x_{k})|\leq n$
for each $k$, contradicting to $|f(x)|\rightarrow\infty$ as $x\rightarrow x_{0}$.
By Bolzano-Weierstrass Theorem, $A_{n}$ is at most countable. (For,
if $A_{n}$ is uncountable, then $A_{n}=\cup_{k\in\mathbb{Z}}\left(A_{n}\cap[k,k+1]\right)$
which implies that $A_{n}\cap[k,k+1]$ is uncountable for some $k\in\mathbb{Z}$.
$A_{n}\cap[k,k+1]$ is a bounded, infinite subset of $\mathbb{R}$,
so it must has accumulation points. A contradiction!). Finally, $\mathbb{R}=\cup_{n}A_{n}$,
contradicting to the fact that $\mathbb{R}$ is uncountable.
