Here is a proposal for a simplim%signum function. Comments?
Return a nounform when nothing known about the limit point:
(%i2) limit(signum(x),x,a);
(%o2) 'limit(signum(x+a),x,0)
Returning a conditional expression instead of a nounform isn't all
that much of a simplification, I think.
When Maxima knows that the limit point is nonzero, use substitution to find
the limit
(%i3) limit(signum(x),x,1/a);
(%o3) signum(1/a)
(%i4) limit(signum(x + a),x,inf);
(%o4) 1
Harmless, I think:
(%i5) limit(signum(x),x,%i);
(%o5) signum(%i)
Two-sided limit toward 0 is undefined (the one-sided limits are defined)
(%i8) limit(signum(x),x,0);
(%o8) und
Bug due do limit(a*x,x,inf) --> infinity
(%i6) limit(signum(a*x+b),x,inf);
(%o6) signum(infinity)
The code
(setf (get '%signum 'simplim%function) 'simplim%signum)
(defun simplim%signum (e x pt)
(let* ((e (limit (cadr e) x pt 'think)) (sgn (mnqp e 0)))
(cond ((eq t sgn) (take '(%signum) e)) ;; limit of argument of signum
is not zero
((eq nil sgn) '$und) ;; limit of argument of signum is zero
(noncontinuous)
(t (throw 'limit nil))))) ;; don't know
Also, I've appended the following to rtest_signum:
limit(signum(x),x,minf);
-1$
limit(signum(x),x,0,'minus);
-1$
limit(signum(x),x,0);
und$
limit(signum(x),x,0,'plus);
1$
limit(signum(x),x,inf);
1$
limit(x * signum(x),x,0);
0$
limit(signum(x+a),x,minf);
-1$
limit(signum(x+a),x,inf);
1$
(assume(notequal(a,0)),0);
0$
limit(signum(x),x,a);
signum(a)$
limit(signum(a*x),x,minf);
-signum(a)$
limit(signum(a*x),x,inf);
signum(a)$
limit(signum(x),x,1/a);
signum(1/a)$
Barton