Continuity of the map $f:0.a_1a_2a_3\ldots \mapsto 0.0a_10a_20a_3\ldots$ I'll first state the question:

Let $f:[0,1] \to [0,1]$ be a function defined as follows: $f(1)=1$, and if $a=0.a_1a_2a_3\ldots$ is the decimal representation of a (which does not end with a chain of 9's), then $f(a)=0.0a_10a_20a_3\ldots$ . Discuss the continuity of $f$ at $0.392$ .

Progress and observations so far: 
Clearly $f:\sum_{i=0}^{\infty}( \frac{a_i}{10^i}) \mapsto \sum_{i=0}^{\infty}( \frac{a_i}{10^{2i}})$ . Take $a=0.a_1a_2a_3a_4\ldots=0.392$ . If I choose $x=0.x_1x_2x_3\ldots$ from a sufficiently small interval on the right side of $a$, say $[a,0.3929]$, then it can be shown that 
$$|f(x)-f(a)|=f(x)-f(a)=\sum_{i} \frac{x_i-a_i}{10^{2i}} \leq \sum_{i} \frac{x_i-a_i}{10^{i}}=|x-a|$$
since for all $i\in \mathbb{N}$ $a_i \leq x_i$ and thus $\frac{x_i-a_i}{10^{2i}} \leq  \frac{x_i-a_i}{10^{i}}$. This proves that $\lim_{x\to a+}f(x)=f(a)$.
The troubles I'm facing: 
I'm not very sure about the above method and I can't think on a better one. Also, by the above method I can't prove $\lim_{x\to a-}f(x)=f(a)$. As the main question uses the word "discuss", I'm not sure whether it really asks to prove or disprove the continuity of $f$ at $a$.
My questions are:

1) Is the function $f$ continuous at $a=0.392$?
2) What can be said about the continuity of $f$ at any arbitrary point from $[o,1]$ rather than the particular point $a=0.392$?
3) Is there any way to view this map graphically?

 A: $f$ is discontinuous at those nonzero numbers whose decimal expansion is finite (i.e. it is of the form $a/10^k$ for $a,k\in \mathbb{N}$). It is continuous at all other fractions, at zero and at all irrational numbers.
A: Let $a_n = 0.391\underbrace{9 \dots 9}_n$ and $a=0.392$. Then $f(a_n) = 0.030901\underbrace{09\dots09}_n$, and so $f(a_n) \to 0.030901 + \frac{9}{99 \times 10^6}$, but $f(a) = 0.030902 \ne \lim_n f(a_n)$ and $a_n \to a$, so $f$ cannot be continuous.
I imagine continuity at an arbitrary point can be disproved using similar modifications: namely, for $a \in (0,1]$ let $a_n = a - \frac{1}{10^n}$ so that $a_n \to a$, then with any luck you can show using a similar argument that $f(a_n) \not \to f(a)$.
It is quite hard to get a graphical image of what's going on here because of the way the problem is defined... and if my hunch is true, $f$ is nowhere continuous (except perhaps at zero), which makes picturing it even harder!
Edit: As Florian's answer states, it is not discontinuous everywhere, but only at a certain subset of the rational numbers, and it is continuous everywhere else. The easiest method to prove this is that used above: show that if the decimal expansion of some $a \in (0,1]$ terminates then there is a $a_n \to a$ for which $f(a_n) \not \to f(a)$; and that if the decimal expansion of $a \in [0,1]$ doesn't terminate or $a=0$ then for any $a_n \to a$ you have $f(a_n) \to f(a)$. 
