> but that problem reminds me of a silly question I've been wanting to
> ask:
questions are rarely silly!
>
> in the case of something like:
>
> (x + 2) (x + 1)
> ---------------
> x + 1
>
> Maxima will simplify this to x+2. But, as written, doesn't this
> expression have a hole at x=-1? It will do the x+2 thing everywhere
> else - is it OK to simplify because it's a case of zero over zero? I
> thought that didn't evaluate to one mathematically.
You can get this straight very easily:
The expression
> (x + 2) (x + 1)
> ---------------
> x + 1
is defined for (complex) numbers x different from -1. It is undefined for
x=-1.
However, the pole at -1 is a so called 'removable singularity', which
means that there is a unique way to define it for x=-1 so that the result
is continuus at x=-1.
For example
f:C->C, x |-> (x+2) for x#-1
x |-> -25*%E*%I for x=-1
is a also an extension of the 'expression' above to all of C but it is not
continuus. So in most cases, it will not be appropriate...
Martin