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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

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bursejan@gmail.com

Posts: 5,239
Registered: 9/25/16
Re: A limit at a point means a function is continuous at that point,
Posted: Nov 10, 2017 11:55 AM
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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 UTC-5, George Cornelius wrote:
> > In article <*-6e5c-470b-bc01-e06cfe543a20@googlegroups.com>,
> > J*hn G*briel <th*n*wcalculus@gmail.com> writes:

> > > On Friday, 10 November 2017 00:15:46 UTC-5, 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 non-number
> > > 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 anal-ysis.
>

> > 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 anal-ysis ?
>

> >
> > Within complex analysis, "at infinity" clearly has a meaning.

>
> Sure. Unicorns have meaning too...
>

> > You deny complex analysis.
>
> I deny all ill-formed concepts and theory derived therefrom. The Great John Gabriel subscribes only to well-formed ideas:
>
> https://www.linkedin.com/pulse/what-does-mean-concept-well-defined-john-gabriel
>

> > 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 Anal-ISIS.
>
> 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.





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