On 12/19/05, Bill Kuncicky <kuncicky@gmail.com> wrote:

> I've gotten the preliminary stuff out of the way (functions, graphs,
> trig, exponentials and logs) and started on the differential calculus
> part.  The first section there is "limits," and I've hit a snag in my
> thinking.  Given a function f(x), such that f(x) = 1 for x .GE. 0, and
> f{x} = 0 for x .LT. 0, is that a continuous function or a discontinuous
> function?
>
> Since for every possible real number x there is a number f(x) which is
> either zero or one, it seems to be that it would be continuous.  The
> book says it is discontinuous, though.  Since they also say "A more
> picturesque description of a continuous function is  that it is a
> function you can graph without lifting your pencil from the paper in the
> region of interest," though, I can see where it might possibly be
> discontinuous, but it is against intuition.

It is discontinuous.

One of the fundamental ideas of calculus is looking at the change in
a function for some small value, call it epsilon, as epsilon is allowed
to approach 0.

So, look at your function f():
   f(0) = 1
and
   f(0-epsilon) = 0
as epsilon approaches 0, f(0-epsilon) does not approach f(0).

Also, again using the idea of epsilon, the derivative of the function f()
at 0 is the limit of
   f(0) - f(0 - epsilon) / epsilon
as epsilon approaches 0.  This is not defined, therefore there is no
derivative of f() at 0, and the function f() is discontinuous at 0.

>  I think that possibly they meant to write  "f(x) = 1 for x .GT  0"
> rather than using the .GE.

No, if they had defined f() such that f(x) = 1 for x GT 0 and
f(x) = 0 for x LT 0, then f() would not be defined at 0, and nothing
could be said about it's derivative at 0 (and thus, nothing about
the functions discontinuity at 0).

Bill  { who seems to be a much bigger geek than he rembered }

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