## Example 6.2.8(a):

It helps to look at the graph of the function:*x*-axis gets smaller and smaller. That indicates that the function is

**not**uniformly continuous - but it is of course not a proof. So:

Take any * = 1*. Does there exist
any such that

if| t - s | <then| f(t) - f(s) | < = 1?

The basic idea is easy: since *| f(t) - f(s) | = | 1/t - 1/s | = | s - t | * 1 / | st |*,
we can see that this might approach infinity if *s* and *t* approach
zero, and therefore will be bigger than any chosen .
All we have to do now is formalize this proof.

Assume there exists such a .
Without loss of generality we may assume that
* < 1* (why ?). Then let
*s = t + / 2*
and set
*t = / 2*.
We have (assuming that *s*, *t* are positive):

| f(t) - f(s) | = | 1/s - 1/t | = | (t - s) / st | = > 1

But now, no matter what * < 1* is we can
make *| f(t) - f(s) | > 1*. Therefore, the function is not uniformly continuous.

This proof, loosely speaking, depends on the fact that after simplification
*| f(t) - f(s) |* goes to infinity if *s* and *t* approach
zero. That is exactly the situation as described in the above picture.