Someone asked me this question. And he said it's an exercise from Rudin's Real and Complex Analysis.
Does there exist a sequence of continuous functions $f_n(x)$, such that $\lim_{n \to \infty} f_n(x)=+\infty$ iff $x \in \mathbb Q$ (or irrationals)?
On the one hand, we know that if the limit of this function exists, then we can't have both as Baire's category theorem applies. But maybe it will happen that the suplim of this sequence is infinity at other points(at which the limit doesn't exists). Because lim equals infinity is the same as inflim equals infinity, so if someone can prove that rationals can't be the (countable)intersections of (countable)unions of G delta set then it's done.
But on the other hand, I suspect that if we let $f_n(x)=\cos(\pi\cdot n!x)^n \cdot n$ is an example.
Best Answer
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$.