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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

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bursejan@gmail.com

Posts: 5,208
Registered: 9/25/16
Re: A limit at a point means a function is continuous at that point,
Posted: Nov 10, 2017 12:24 PM
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Come on bird brain John Garbage-iel, 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 super-idiot 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 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 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 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!





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