Prove that the function f defined by $\displaystyle f(x)= x $if is rational and $\displaystyle f(x)=-x$ if x is irrational is continuous at 0 only .

Ok so I was thinking doing it this way:

$\displaystyle \exists \{x_n\} \in \mathbb{Q}$ such that $\displaystyle x_n \to c$, $\displaystyle x_n \neq c $, $\displaystyle \forall n$ and $\displaystyle c \neq 0$

$\displaystyle \exists \{y_n\} \in \mathbb{R} \backslash \mathbb{Q}$ such that $\displaystyle y_n \to c$ ,$\displaystyle y_n \neq c $ ,$\displaystyle \forall n$ and $\displaystyle c \neq 0$

Then,

$\displaystyle lim f(x_n) = lim (x_n)=c$

$\displaystyle

lim f(y_n) = lim -(y_n)=-c

$

So then if I say something like there exists a sequence $\displaystyle \{x_n\} $ of rational numbers such that $\displaystyle x_n \to 0$ and a sequence $\displaystyle \{y_n\} $ of irrational number such that $\displaystyle y_n \to 0$. And then show that there limits are the same would that show that it is only continuous at 0???