Consider the function f with f(x) = 1 if x is rational and f(x) = 0 is x is irrational. Does the limit of f(x) exist at an arbitrary number x ?Recall that f is called the Dirichlet function. Since f 'jumps wildly' up and down, we suspect that the function does not have a limit at any point. This is indeed the case, as is easy to show using our new definition of limit:
Let c be any real number and pick = 1/2. Suppose there was a > 0 such that whenever | x - c | < then | f(x) - L | < = 1/2.
We can find a rational number q with | q - c | < . Hence f(q) = 1, and therefore
| f(q) - L | < 1/2, or | 1 - L | < 1/2or equivalently
| L | > 1/2We can also find an irrational number y with | y - c | < . Hence f(y) = 0, and therefore
| f(y) - L | < 1/2or
| L | < 1/2But that's a contradiction, so the function can not have a limit at any point c.