12/20/2007, 09:50 PM

In my notation, it would be (as infinity is 1/i for grade i infinitesimals):

f(1/i) = ln(1/i)*(1/i)^2 = -lni* (-i)^2 = (-i)^3 * pi/2

So its big oh limit would be 3, but pi/2 would appear. This means this limit will be transcending the squared infinity because of extra -i*pi/2. What does it mean to transcend infinity that is created by 1/infinitesimal I do not know yet.

Maybe it is like exausting infinity, since pi continues beyond any infinity in length.... --i*pi/2 is obviously not a usual infinitesimal just multiplied by a finite real because -i^2 * e^-ipi/2 = -k where -k is infinity perpendicular to -i ^2 (not in complex plane, but perpendicular to it).

I would say big oh limit for this function is 2 + (1 turned by 90 degrees). At complex plane, it may look like 2+ infinitesimal, but that would be just a projection of -k on complex plane ( the 3rd i than is infinitesimal multiplied by -pi/2).

Well, not exactly like this, of course....

f(1/i) = ln(1/i)*(1/i)^2 = -lni* (-i)^2 = (-i)^3 * pi/2

So its big oh limit would be 3, but pi/2 would appear. This means this limit will be transcending the squared infinity because of extra -i*pi/2. What does it mean to transcend infinity that is created by 1/infinitesimal I do not know yet.

Maybe it is like exausting infinity, since pi continues beyond any infinity in length.... --i*pi/2 is obviously not a usual infinitesimal just multiplied by a finite real because -i^2 * e^-ipi/2 = -k where -k is infinity perpendicular to -i ^2 (not in complex plane, but perpendicular to it).

I would say big oh limit for this function is 2 + (1 turned by 90 degrees). At complex plane, it may look like 2+ infinitesimal, but that would be just a projection of -k on complex plane ( the 3rd i than is infinitesimal multiplied by -pi/2).

Well, not exactly like this, of course....