Answer
Please see below.
Work Step by Step
According to the definition of continuity of a function, a necessary condition for a function to be continuous at a point is that the limit of the function does exist at the point. Now, we claim that the Dirichlet function does not have limit at any real number.
We prove the claim by contradiction. Suppose that there exists some real number $c$ such that the Dirichlet function has limit $L$ at this point. So for $\varepsilon =\frac{1}{2}$ there must exist a $\delta _0$ such that$$|x-c|< \delta _0 \quad \Rightarrow \quad |f(x)-L| < \frac{1}{2}.$$As we know, between any two real numbers there exist rational and irrational numbers, so in any interval there exist a rational and irrational numbers within the interval. So let us choose some rational number $y$ and irrational number $z$ within the interval $(c - \delta _0 , c+\delta _0 )$. According to the definition of the Dirichlet function, we can get the following by substituting them into the above inequality:$$|f(y)-L|=|0-L|=|L|
