Solutions

Suppose

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

for contradiction. By definition, there exists a $\delta$ such that

$$0<|x-a|<\delta \Rightarrow |f(x)-l|<1.$$

However, because both rational and irrational numbers are dense in the reals, the implied inequality does not hold no matter what value of $l$ is chosen.