Since there is a homeomorphism of $\Bbb R $ taking $\Bbb Q $ to the dyadic rationals, that is, rational numbers where the denominator is a power of 2, (see here for an example of such a function), we can restrict out attention to finding a sequence of functions converging to infinity exactly at the dyadics.
Now for $ p, d \in \Bbb R $ we let $ h_{p, d} $ denote a hat function, i. e. a continuous function with $ h_{p, d}(p) =1$ and $ h_{p, d}(x) =0$ for $ x \notin [p-d, p+d]$.
Now define
$ f_n = \sum_{k \in \Bbb Z} n h_{k/2^n,2^{-n-2} }$
It is obvious that for dyadic x, $ f_n (x) \to \infty $ as $ n \to \infty $.
If $x$ is not dyadic, there are infinitely many $ n $ with $\{2^n x\} \in [1/4,1/2]$ (here the brackets denote the fractional part of the real number). To see this, consider the binary expansion of $ x $ and note that, since this expansion is not allowed to end in an infinite chain of zeros or in an infinite chain of ones, there are infinitely many integers $ n $ such that the $ n+1$-th digit after the dot is a $0$ and the $ n+2$-th is a $1$.
But with such a choice of $ n $, we get $ f_n (x)=0$ which shows that $ f_n (x )\not \to \infty $ as $ n \to \infty$.
To show that $E$ is open: Let $F=E^{c}=\{x\mid\limsup_{y\rightarrow x}|f'(y)|=\infty\}$. We
go to show that $F$ is closed. Let $(x_{n})$ be a sequence in $F$
and suppose that $x_{n}\rightarrow x$ for some $x\in\mathbb{R}$.
We prove that $x\in F$ by contradiction. Suppose the contrary that
$x\notin F$. Choose $M>0$ such that $\limsup_{y\rightarrow x}|f'(y)|<M$.
There exists $\delta>0$ such that $|f'(y)|<M$ whenever $y\in(x-\delta,x+\delta)\setminus\{x\}$.
Since $x_{n}\rightarrow x$ and $x_{n}\neq x$, there exists $n$
such that $x_{n}\in(x-\delta,x+\delta)\setminus\{x\}$. Without loss
of generality, we suppose that $x<x_{n}<x+\delta$. Choose $\varepsilon>0$
be sufficiently small such that $x<x_{n}-\varepsilon<x_{n}<x_{n}+\varepsilon<x+\delta$.
Since $\lim_{y\rightarrow x_{n}}|f'(y)|=\infty$, there exists $y_{0}\in(x_{n}-\varepsilon,x_{n}+\varepsilon)\setminus\{x_{n}\}$
such that $|f'(y_{0})|>2M$. Note that $y_{0}\in(x-\delta,x+\delta)\setminus\{x\}$,
so we also have $|f'(y_{0})|<M$, which is a contradiction.
In the above, we have not used any properties about $f$ nor its derivative
$f'$. That is, that $F$ is closed continues to hold if $f'$ is
replaced by an arbitrary function $g:\mathbb{R}\rightarrow\mathbb{R}$.
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To show that $E$ is dense: Note that $E^{-}=\mathbb{R}$ iff $\emptyset=E^{-c}=\left(E^{coc}\right)^{c}=F^{o}$.
That is, we need to show that $F$ has empty interior. Prove by contradiction.
Suppose that there exist $\alpha<\beta$ such that $(\alpha,\beta)\subseteq F$.
For each $n\in\mathbb{N}$, let $A_{n}=\{x\in(\alpha,\beta)\mid f'(x)\in(-n,n)\}$,
which is a $F_{\sigma}$-subset of the topological space $(\alpha,\beta)$
(See Theorem 2 in the appendix). For each $n$, write $A_{n}=\cup_{k}F_{nk}$,
for some closed subsets $F_{nk}$ of $(\alpha,\beta)$. Note that
$(\alpha,\beta)=\cup_{n} A_n=\cup\{F_{nk}\mid n,k\in\mathbb{N}\}$. Since
$(\alpha,\beta)$ is a Baire space, by Baire Category Theorem, there
exist $n,k$ such that $F_{nk}$ has non-empty interior. That is,
there exist $\alpha'<\beta'$ such that $(\alpha',\beta')\subseteq F_{nk}\subseteq A_{n}$.
Choose $x_{0}\in(\alpha',\beta')$. Note that $x_{0}\in F$, so there
exists a sequence $(x_{k})$ with $x_{k}\neq x_{0}$, $x_{k}\rightarrow x_{0}$,
and $|f'(x_{k})|\rightarrow\infty$. Observe that $x_{k}\in(\alpha',\beta')$
for large $k$ and hence $|f'(x_{k})|<n$, contradicting to $|f'(x_{k})|\rightarrow\infty$.
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Appendix. We state and prove the following theorems.
Theorem 1: Let $X$ be a topological space. Let $f_{n}:X\rightarrow\mathbb{R}$
and $f:X\rightarrow\mathbb{R}$. Suppose that $f_{n}$ is continuous
and $f_{n}(x)\rightarrow f(x)$ for each $x\in X$. Then, for each
open subset $O\subseteq\mathbb{R}$, $f^{-1}(O)$ is a $F_{\sigma}$-subset
of $X$ (i.e., countable union of closed subsets).
Proof of Theorem 1: See my other post $f_n\rightarrow f$ pointwise, $O$ open subset of $\mathbb{R}$ $\Rightarrow$ $f^{-1}(O)$ is $F_{\sigma}$
Theorem 2: Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a differentiable
function. If $O\subseteq\mathbb{R}$ is open, then $f'^{-1}(O)$ is
a $F_{\sigma}$-set.
Proof of Theorem 2: For each $n$, let $f_{n}:\mathbb{R}\rightarrow\mathbb{R}$
be defined by $f_{n}(x)=n\left[f(x+\frac{1}{n})-f(x)\right]$. Note
that $f_{n}$ is continuous and $f_{n}(x)\rightarrow f'(x)$ for each
$x\in\mathbb{R}$. Now, the result follows from Theorem 1.
Best Answer
Let $f_n(x)$ denote the function whose graph is an isosceles triangle with base $[0,1/n]$ and height $1$. Specifically: $$f_n(x) = \begin{cases} 2nx & \text{if }0 \leq x \leq \frac{1}{2n} \\ 2 - 2nx & \text{if }\frac{1}{2n} \leq x \leq \frac{1}{n} \\ 0 & \text{otherwise} \end{cases}$$ and let $$f(x) = \sum_{n=1}^{\infty}f_n(x-n)$$ So, the graph of $f$ consists of a series of increasingly narrow triangles, where the $n$'th triangle has base $[n, n+1/n]$ and height $1$. It's straightforward to verify that $$\lim_{n \to \infty}f(x+n) = 0$$ for all $x$, but $\lim_{x \to \infty}f(x)$ does not exist.