# Interactive Real Analysis - part of MathCS.org

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## Example 6.1.5:

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 | < 1/2
or equivalently
| L | > 1/2
We can also find an irrational number y with | y - c | < . Hence f(y) = 0, and therefore
| f(y) - L | < 1/2
or
| L | < 1/2
But that's a contradiction, so the function can not have a limit at any point c.
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