Is there a function $f: \mathbb{R} \to \mathbb{R}$ such that $\lim\limits_{x\to p}f(x)=\infty$ for every $p \in \mathbb{R}$ 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.
 A: If $f$ is such a function and $M>0$ then 
$$ \forall x_0\in \! [0,1]  \qquad \exists \delta_{x_0}\ \qquad \ 0<|x-x_0|<\delta_{x_0} \implies f(x)>M.$$
But
$$\bigcup_{x_0\in [0,1]} B(x_0,\delta_{x_0})\setminus \{x_0\}\supseteq [0,1]$$
and $[0,1]$ is compact So there exists $x_1,\ldots , x_n$ such that 
$$B(x_1,\delta_{x_1})\cup \ldots \cup B(x_n,\delta_{x_n}) \setminus \{x_1,\dotsc,x_n\}\supseteq [0,1]$$
So $f(x)>M$ for all $x\in [0,1]\setminus\{x_1,\ldots,x_n  \} $.
So $ \{ x: f(x)\le M \} $ is at most countable and $$\bigcup\limits_{M\in \mathbb{N}} \{ x:f(x)\le M \}$$ is at most countable. Since $[0,1]$ is uncountable, this is absurd hence no such $f$ exists.
A: 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 $\limsup\limits_{k\to\infty}|f(x_k)|\leqslant n$.
