real analysis - Discontinuity of second kind with some properties

My friend ask me to construct a function with infinite discontinuity of second kind(i.e. one of $\lim_{x\to x_0^-}f(x)$ and $\lim_{x\to x_0^+}f(x)$ doesn't exists) defined on $[0,1]$, such that the rational numbers are discontinuity of second kind and the irrational numbers are all continuous.

I have some idea of constructing it,but not sure if this is right. Consider the mapping $E=\{1,1/2,1/4...1/2^n...\}$, $g:\mathbb{Q}\to E$ and $g$ is a bijection. We define $f$ to be
$$f(x)=\sum_{q_i\in\mathbb{Q},q_iBut I can't prove that if this has the property stated above.Thanks for your attention & advice.

Answer

First, the "$q_i

$f(x)=\displaystyle\sum_{q_i\in\mathbb{Q}}\frac{1}{2^i}\sin\left(\frac{1}{x-q_i}\right)$.

Fix some rational $q_M\in[0,1]$. Then write $f(x)=f_M(x)+\widehat{f_M}(x)$, where $f_M(x)=\frac{1}{2^M}\sin\left(\frac{1}{x-q_M}\right)$, and $\widehat{f_M}(x)=\displaystyle\sum_{q_i\in\mathbb{Q},i\neq M}\frac{1}{2^i}\sin\left(\frac{1}{x-q_i}\right)$.

It is clear that $\lim_{x\to {q_M}^+}f_M(x)$ and $\lim_{x\to {q_M}^-}f_M(x)$ do not exist, so to show that $\lim_{x\to {q_M}^+}f(x)$ and $\lim_{x\to {q_M}^-}f(x)$ also do not exist, it suffices to show that $\widehat{f_M}$ is continuous at $q_M$.

Fix some $\epsilon>0$. Choose some $N>0$ such that $\displaystyle\sum_{i=N}^{\infty}\frac{1}{2^i}<\frac{\epsilon}{2}$. Now choose some $\delta>0$ such that if $|x-q_M|<\delta$,

$\displaystyle
\left|\sum_{i=1}^{N-1}\frac{1}{2^i}\sin\left(\frac{1}{q_M-q_i}\right)-\sum_{i=1}^{N-1}\frac{1}{2^i}\sin\left(\frac{1}{x-q_i}\right)\right|<\frac{\epsilon}{2}.

$

Then for $|x-q_M|<\delta$, $|\widehat{f_M}(x)-\widehat{f_M}(q_M)|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$, so $\widehat{f_M}$ is continuous at $q_M$. Therefore we conclude that neither $\lim_{x\to {q_M}^+}f(x)$ nor $\lim_{x\to {q_M}^-}f(x)$ exist.

(Note that an almost identical argument will show that $f$ is continuous at each non-rational point in $[0,1]$.)