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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.

If $f$ is such a function and $M>0$ then
$$ forall x_0in ! [0,1] qquad exists delta_{x_0} qquad 0<|x-x_0|<delta_{x_0} implies f(x)>M.$$
But
$$bigcup_{x_0in [0,1]} B(x_0,delta_{x_0})setminus {x_0}supseteq [0,1]$$
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 $xin [0,1]setminus{x_1,ldots,x_n } $.
So $ { x: f(x)le M } $ is at most countable and $$bigcuplimits_{Min mathbb{N}} { x:f(x)le M }$$ is at most countable. Since $[0,1]$ is uncountable, this is absurd hence no such $f$ exists.

Let us show that such a function $f$ does not exist.

Choose any function $f:mathbb Rtomathbb R$ and, for every positive integer $n$, consider the set $$A_n={xin[0,1],;,|f(x)|leqslant n}$$ 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 $x_kne x_j$ for every $kne j$ and $x_k$ is in $A_n$ for every $k$. Then the set $X={x_k,;,kgeqslant0}$ is infinite and such that $Xsubseteq A_n$. Since $[0,1]$ is compact, $X$ has a limit point $p$ in $[0,1]$. Assume without loss of generality that $limlimits_{ktoinfty}x_k=p$ and that $x_kne p$ for every $k$.

Now, $|f(x_k)|leqslant n$ for every $k$ hence $limlimits_{xto p}|f(x)|=+infty$ is impossible because $limlimits_{ktoinfty}x_k=p$ with $x_kne p$ for every $k$ and $limsuplimits_{ktoinfty}|f(x_k)|leqslant n$.