[NBLUG/talk] OT division by zero OT

[Andru Luvisi]

On Sat, 31 May 2003, Steve Zimmerman wrote:
[snip]
> Steve2:  Yes, but then the graph of 1/x is discontinuous, and therefore
>                not a function, and if not a function, then not amenable to the
> 	 limit theorem.
>
> Steve1:  Correct.  We are taught in calculus that the limit theorem only
> 	 works on functions, and functions are, by definition,
> 	 continuous.
[snip]

I thought I was done with this thread, but you just gave me a chance to
show off one of my favorite functions, and who can resist that?

As it happens, a function can be continuous without approaching a limit
and can approach a limit without being continuous.

Let g(x) = sin(1/x)

g(x) is defined and continuous for all x > 0, but g(x) does not approach
any limit as x approaches 0.  As a matter of fact, for all n > 0, g(x)
oscillates a countably infinite nuber of times between 1 and -1 over the
open interval (0, n).

Let f(x) be:
    x*g(x)     if g(x) >= 0
    x*g(x) + x if g(x) < 0

For all n > 0, f(x) has a countably infinite number of discontinuities
over the open interval (0, n), but it is trivial to prove that f(x)
approaches 0 as x approaches 0.  Just let delta equal epsilon and note
that 0 <= f(x) <= x for all x > 0.

Andru
-- 
Andru Luvisi, Programmer/Analyst

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