# Thread: Showing a function divereges at C

1. ## Showing a function divereges at C

$f: R->R$ is defined by $f(x)=x$ if $x$ is rational, and $f(x)=0$ is irrational.

I need to show f does not have a limit at c if c is not equal to 0. I've shown this true if c is rational, but I'm having a problem with if it's irrational, any help?

2. Suppose that $\varepsilon > 0\,\& \,c \ne 0$ then due to density
$\left( {\exists q \in \mathbb{Q}} \right)\left( {\exists i \in \Re \backslash \mathbb{Q}} \right)\left[ {\left\{ {q,i} \right\} \subseteq \left( {c - \varepsilon ,c + \varepsilon } \right)} \right]$.
Thus in any epsilon neighborhood of c there is a rational and an irrational number.
Thus on that neighborhood of a nonzero c the function is both c & 0.