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Topic:
A limit at a point means a function is continuous at that point,
Replies:
4
Last Post:
Nov 11, 2017 8:27 AM




Re: A limit at a point means a function is continuous at that point,
Posted:
Nov 10, 2017 12:24 PM


Come on bird brain John Garbageiel, you know you are no saint. Only since you can control yourself
today, doesn't mean you can control yourself today. So far there is usually a spamming stream by a moron
and superidiot bird brain John Gabriel, who denies all kind of stuff such as:  real numbers  limit  infinite series  piecewise functions  etc..
But nevertheless has a new calculoose, which even cannot differentiate f(x)=x=y.
Am Freitag, 10. November 2017 18:08:31 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 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 don't > > really care about this ongoing discussion in which you claim > > to have overturned all mathematics as we know it. > > > > So let's agree to something. I gave a solution which is within > > the orthodox mathematical views. That is all. To those who > > do not accept such views, my post is of no interest. It is > > undefined in your world. > > > > 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. > > > > Within complex analysis, "at infinity" clearly has a meaning. You > > deny complex analysis. We're on different playing fields. 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. > > > > 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. > > See what I mean Mr. Cornelius? Jan Burse produced 3 new turds on my thread since you fed him. Please do not feed the troll  especially in my threads. Thank you in arrear and in advance!



