* Raymond Toy <rtoy@earthlink.net> [2005-09-16 12:34:18 -0400]:
>
> >
> > I am puzzled by the introduction of log(-1) in the result. The
> > problem in the bug is lim(f/g) type problem, where both lim(f) and
> > lim(g) are finite, lim(g) # 0. I am wondering if Maxima's limit
> > algorithm does include treating that kind of basic case, and if so,
> > does Maxima "know" that limit(atan(x),x,inf)=%pi/2, and that
> > limit(exp(-t),t,inf)=0?
>
> I thought the same, but for some reason, maxima didn't try f/g. But
> clearly maxima does know the limits for atan and exp.
>
How difficult would it be to modify the Limits package to make sure
that Maxima looks for the "easiest" possibilities first, such as
checking lim(f), lim(g) and if those exists, if lim(g)#0?
> >
> > Back to the bug: For problems like one in the bug, I would like to
> > believe that Maxima ought to be able to handle those staying in the
> > domain of real numbers, and do "simple" evaluation lim(f)/lim(g).
> > Perhaps yours "add %log to one of the special cases in simplimit
> > (clause 3)." is going to do exactly that. I apologize if I missed that
>
> Unfortunately, no, that doesn't do that. What it does is prevent maxima
> from doing lim(f(x)) = f(lim x) for finite x. This is wrong of course
> if f(x) has discontinuities for finite values. I also don't understand
> why sin, cos, and others are listed here.
>
Thanks for the clarification.
> There are probably other possible approaches for the issue in bug
> 1281737. The fix I made is really meant for the log bug, which
> concidentally fixes bug 1281737.
>
Nobody is going to complain for getting two birds by one stone :)
> > However, if Maxima has to run into limit(log(w(z)),z,a) type of
> > thing, the it seems that the lim(log) =log(lim) does not hold if the
> > limit point happens to be in the negative real axis. Your approach to
> > evaluate limits of both real and imaginary part seem to be a good way
> > to go, with a proviso - the sign of the imaginary part should be kept
> > constant to ensure approach to the limit point either through the
> > second or the third quadrant.
>
> I believe it does preserve the sign.
>
Are you sure about this? It seems to me that if Maxima says that
log(-1)=%i*%pi, and if it keeps the track of the sign of the imaginary
part in log(1-%i*x), then it should not return log(-1) for the
limit(log(-1-%i*x),x,0,plus).
Milan
> Ray
>
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