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Topic: A limit at a point means a function is continuous at that point,
Replies: 4   Last Post: Nov 10, 2017 11:51 AM

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 bursejan@gmail.com Posts: 5,511 Registered: 9/25/16
Re: A limit at a point means a function is continuous at that point,
Posted: Nov 10, 2017 10:02 AM

L'Hopital doesn't fail for:

f(b) - f(a)
----------- = ? for lim b->a
b - a

Take a as a constant, then f(a) is also a constant.
The derivatiev of a constant is 0, so we have:

d a/db = 0

d f(a)/db = 0

Now take the derivative of the numerator and denumerator,
you get the following:

d(f(b)-f(a))/db f'(b) - 0
--------------- = --------- = f'(b) for lim b->a
d(b-a)/db 1 - 0

So we can find the limit of this zero by zero division,
using L'Hopital rule, and we get:

f(b) - f(a)
----------- = f'(b) for lim b->a
b - a

Exercise: Use chain rule and other movements
of a,b, what will be the result of limit?

Am Freitag, 10. November 2017 13:36:24 UTC+1 schrieb John Gabriel:
> 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.
>

> >
> > 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.
>

> > 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.
> >

> > > Am Donnerstag, 9. November 2017 14:42:37 UTC+1 schrieb Zelos Malum:
> > >> >The idiotic mainstream tendency is to transfer the limit to the exponent, that is,
> > >>
> > >> They do not
> > >>

> > >> >Then what do orangutans do? They say:
> > >> >
> > >> >0=e^(-oo)

> > >>
> > >> Nope, for real numbers that is not defined or meaningful so they don't do that. Iti s your idiocy not theirs.

Date Subject Author
11/10/17 bursejan@gmail.com
11/10/17 bursejan@gmail.com
11/10/17 bursejan@gmail.com
11/10/17 George Cornelius