Answer
$g$ is continuous at $x=0$
Work Step by Step
$g(x) = 0~~$ if $x$ is rational
$g(x) = x~~$ if $x$ is irrational
We can show that $g$ is continuous at $0$
Let $\epsilon \gt 0$ be given.
Let $\delta = \epsilon$
Suppose that $0 \lt \vert x-0 \vert \lt \delta$
Then:
If $x$ is rational, $\vert g(x)-g(0) \vert =\vert 0-0 \vert =0 \lt \epsilon$
If $x$ is irrational, $\vert g(x)-g(0) \vert =\vert x-0 \vert \lt \delta = \epsilon$
Therefore, $g$ is continuous at $0$
