TITLE: Is there a function $f: \mathbb{R} \to \mathbb{R}$ such that $\lim\limits_{x\to p}f(x)=\infty$ for every $p \in \mathbb{R}$
QUESTION [6 upvotes]: I was wondering if there is a function $f: \mathbb{R} \to \mathbb{R}$ such that its limit at every point is infinite. 
I guess not, because what would its graph look like, but then again, I don't know how to prove it.

REPLY [5 votes]: Let us show that such a function $f$ does not exist.
Choose any function $f:\mathbb R\to\mathbb R$ and, for every positive integer $n$, consider the set $$A_n=\{x\in[0,1]\,;\,|f(x)|\leqslant n\}$$ then, since $f(x)$ is finite for every $x$, $$\bigcup_nA_n=[0,1]$$ in particular, there exists some $n$ such that $A_n$ is infinite.
Pick a sequence $(x_k)$ such that $x_k\ne x_j$ for every $k\ne j$ and $x_k$ is in $A_n$ for every $k$. Then the set $X=\{x_k\,;\,k\geqslant0\}$ is infinite and such that $X\subseteq A_n$. Since $[0,1]$ is compact, $X$ has a limit point $p$ in $[0,1]$. Assume without loss of generality that $\lim\limits_{k\to\infty}x_k=p$ and that $x_k\ne p$ for every $k$.
Now, $|f(x_k)|\leqslant n$ for every $k$ hence $\lim\limits_{x\to p}|f(x)|=+\infty$ is impossible because $\lim\limits_{k\to\infty}x_k=p$ with $x_k\ne p$ for every $k$ and $\limsup\limits_{k\to\infty}|f(x_k)|\leqslant n$.