Solutions

Suppose for contradiction that, for $a\neq 0$,

$$\lim_{x\to a} f(x)=l.$$

Let

$$\varepsilon=\min(|l-a|,|l+a|).$$

By definition, there exists $\delta$ such that

$$0<|x-a|<\delta \Rightarrow |f(x)-l|<\varepsilon=\min(|l-a|,|l+a|).$$

Again, because both rationals and irrationals are dense in the reals, the implied inequality does not hold for any value of $l$.