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Topic:
A limit at a point means a function is continuous at that point,
Replies:
3
Last Post:
Nov 12, 2017 1:45 PM




Re: A limit at a point means a function is continuous at that point,
Posted:
Nov 10, 2017 11:55 AM


Dont worry bird brain John Gabriel, we will help you with analysis. Don't be affraid to ask.
Am Freitag, 10. November 2017 17:53:17 UTC+1 schrieb John Gabriel: > On Friday, 10 November 2017 10:57:49 UTC5, George Cornelius wrote: > > In article <*6e5c470bbc01e06cfe543a20@googlegroups.com>, > > J*hn G*briel <th*n*wcalculus@gmail.com> writes: > > > On Friday, 10 November 2017 00:15:46 UTC5, George Cornelius wrote: > > > > > >> > Somewhere else I wrote 2*e^(1/x^2)/x^3 = 1. Sorry for > > >> > that error, but in fact it is also: > > >> > > > >> > lim x>0 2*e^(1/x^2)/x^3 = 0 > > >> > > >> > Which is a nice little limit exercise. > > >> > > >> The exponential function has an "essential singularity" at infinity. > > > > > > There is nothing "essential" in mathematics and "at infinity" is > > > an impossible scenario. We simply say the function is undefined at > > > x=0 and avoid use of words such as singularity which morons like > > > Stephen Hawking have overused. > > > > You have never taken a course in complex analysis, have you? > > I've done better  I have compiled courses in complex analysis. > > > > > >> I have long forgotten my complex analysis but was going to guess that > > >> that implied that it grows near infinity faster than any power > > >> of x, and thus x^(3)/e^(1/x^2) goes to zero as x goes to zero. > > > > > > There is no limit as I have proved. You cannot substitute a nonnumber > > > that is, oo into e^x. e^(oo) is meaningless nonsense. oo is not a limit. > > > > > >> > > >> But what do I know? L'Hopital is a bit tricky, but using > > > > > > You can't use L'Hopital's property unless it has the form oo/oo > > > or 0/0 which is not case here. Also L'Hopital fails for several > > > functions which do have those forms. > > > > If you don't mind, I was responding to someone else. > > I'm sorry! You mean the new spamming troll Jan Burse? He doesn't know any mathematics, never mind complex analysis. > > > I don't really care about this ongoing discussion in which you claim > > to have overturned all mathematics as we know it. > > I do! > > > > > So let's agree to something. I gave a solution which is within > > the orthodox mathematical views. > > But that is not the topic of this thread. Encouraging the troll Jan Burse is not good for this thread in which he has already contributed a considerable amount of diarrhea. > > > That is all. To those who > > do not accept such views, my post is of no interest. It is > > undefined in your world. > > It is undefined in any world. You just don't know it. > > > > > It's kind of like playing soccer, or futbol, vs. American > > football. Don't like the game because it's not played by > > your favorite set of rules? Change the channel. > > See, mathematics is not about rules and if you consider this to be a channel, then I have news for you! You are on the wrong channel. Perhaps you were looking for a thread called sci.myth complex analysis ? > > > > > Within complex analysis, "at infinity" clearly has a meaning. > > Sure. Unicorns have meaning too... > > > You deny complex analysis. > > I deny all illformed concepts and theory derived therefrom. The Great John Gabriel subscribes only to wellformed ideas: > > https://www.linkedin.com/pulse/whatdoesmeanconceptwelldefinedjohngabriel > > > We're on different playing fields. > > Come to think of it, we're probably not even on the same planet! :) > > > Heckle the referees on your playing field but don't come running over to > > mine and telling me that I'm playing some game prevalent in the > > universe you inhabit by the wrong rules. > > Whatever ... I started this thread and you Mr. George are clearly gate crashing my party. While you are free to comment on the topic, it's kind of rude to introduce a different topic  Complex AnalISIS. > > Do have a wonderful earth day or night! > > > > > George > > > > >> the limit of the log of the function gives you a straightforward > > >> solution for x>0+ . For the other side of the limit you want > > >> (x)^3/e^(1/x^2) as x>0+, which will be the negative of the > > >> other limit and therefore zero as well.



