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Robert B. Israel science forum Guru
Joined: 24 Mar 2005
Posts: 2151

Posted: Mon Jun 26, 2006 5:29 pm Post subject:
Re: limit(ln(x+sqrt(x^2+1)),x=infinity)



David W. Cantrell wrote:
Quote:  israel@math.ubc.ca (Robert Israel) wrote:
limit(arctanh(x), x=1) returns infinity.
This should be undefined,
I'm not sure that it should be undefined, because I don't know exactly what
extension of C is used by Maple.
as the limit from the left (using Maple's
choice of the branch cut) should be infinity + pi i/2.
That's fine, so far. But then how does that simplify? For example, in the
extension of C used by Mathematica, we have
In[1]:= Infinity + Pi I/2
Out[1]= Infinity
in other words, the finite part was absorbed by the infinite part.
The virtues of various extensions could certainly be debated. But at least
in Mathematica's extension, in a sum of a finite term and an infinite term,
the finite term is always absorbed.

Maple's simplification does not change infinity + pi i/2. It
simplifies real and
imaginary parts separately.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada 

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David W. Cantrell science forum Guru
Joined: 02 May 2005
Posts: 352

Posted: Mon Jun 26, 2006 2:05 pm Post subject:
Re: limit(ln(x+sqrt(x^2+1)),x=infinity)



israel@math.ubc.ca (Robert Israel) wrote:
Quote:  In article <1151266465.658389.154710@i40g2000cwc.googlegroups.com>,
Gene Ward Smith <genewardsmith@gmail.com> wrote:
Fooling around with Maple, I see it takes values in the extended reals,
but won't give them. That is, arctan(infinity) = pi/2 according to
Maple, but tan(pi/2) returns an error message.
Similarly with tanh(infinity) and arctanh(1). We must be
evervigilant, I suppose.
limit(tan(x), x=Pi/2, right) returns infinity, as it should.
limit(arctanh(x), x=1) returns infinity.
This should be undefined,

I'm not sure that it should be undefined, because I don't know exactly what
extension of C is used by Maple.
Quote:  as the limit from the left (using Maple's
choice of the branch cut) should be infinity + pi i/2.

That's fine, so far. But then how does that simplify? For example, in the
extension of C used by Mathematica, we have
In[1]:= Infinity + Pi I/2
Out[1]= Infinity
in other words, the finite part was absorbed by the infinite part.
The virtues of various extensions could certainly be debated. But at least
in Mathematica's extension, in a sum of a finite term and an infinite term,
the finite term is always absorbed.
David W. Cantrell 

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Robert B. Israel science forum Guru
Joined: 24 Mar 2005
Posts: 2151

Posted: Mon Jun 26, 2006 6:27 am Post subject:
Re: limit(ln(x+sqrt(x^2+1)),x=infinity)



In article <1151266465.658389.154710@i40g2000cwc.googlegroups.com>,
Gene Ward Smith <genewardsmith@gmail.com> wrote:
Quote:  Fooling around with Maple, I see it takes values in the extended reals,
but won't give them. That is, arctan(infinity) = pi/2 according to
Maple, but tan(pi/2) returns an error message.
Similarly with tanh(infinity) and arctanh(1). We must be
evervigilant, I suppose.

limit(tan(x), x=Pi/2, right) returns infinity, as it should.
limit(arctanh(x), x=1) returns infinity.
This should be undefined, as the limit from the left (using Maple's
choice of the branch cut) should be infinity + pi i/2.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada 

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Gene Ward Smith science forum Guru
Joined: 08 Jul 2005
Posts: 409

Posted: Sun Jun 25, 2006 11:32 pm Post subject:
Re: limit(ln(x+sqrt(x^2+1)),x=infinity)



David W. Cantrell wrote:
Quote:  And speaking of _abus de langage_, arc<hyperbolic function> is a misnomer,
although often used nowadays.

I know people like to complain about it, but I also know it's what
Maple uses. It's been around a long time, and arctanh is an improvement
over tanh^(1) in the misnomer department. 

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Gene Ward Smith science forum Guru
Joined: 08 Jul 2005
Posts: 409

Posted: Sun Jun 25, 2006 11:29 pm Post subject:
Re: limit(ln(x+sqrt(x^2+1)),x=infinity)



Dirk Van de moortel wrote:
Quote:  "Gene Ward Smith" <genewardsmith@gmail.com> wrote in message news:1151266465.658389.154710@i40g2000cwc.googlegroups.com...
Fooling around with Maple, I see it takes values in the extended reals,
but won't give them. That is, arctan(infinity) = pi/2 according to
Maple, but tan(pi/2) returns an error message.
It should give an error message since the limit does not exist.
Left and right limits are +inf and inf.
I hope Maple doesn't give inf for limit{ x > pi/2 ; tan(x) } ?

That evaluates to "undefined". F(undefine) then always returns
undefined, apparently. 

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David W. Cantrell science forum Guru
Joined: 02 May 2005
Posts: 352

Posted: Sun Jun 25, 2006 9:19 pm Post subject:
Re: limit(ln(x+sqrt(x^2+1)),x=infinity)



"Dirk Van de moortel" <dirkvandemoortel@ThankSNOSperM.hotmail.com> wrote:
Quote:  "Gene Ward Smith" <genewardsmith@gmail.com> wrote in message
news:1151266465.658389.154710@i40g2000cwc.googlegroups.com...
Dirk Van de moortel wrote:
I just don't think the concept of "extended reals" is useful in
real analysis.
My "attains to increasingly negative values etc" shows one use; that's
really too sloppy to be an actual definition. I can say "f(x) 
infinity" iff arctan(f(x)) > pi/2", but then I'm really using the
extended reals and may as well be explicit about it, projecting the
real numbers onto a semicircular arc and sticking in pi/2 and pi/2 as
the endpoints of the arc.
Fooling around with Maple, I see it takes values in the extended reals,
but won't give them.

Unlike Mathematica, which giveth and taketh.
Quote:  That is, arctan(infinity) = pi/2 according to
Maple, but tan(pi/2) returns an error message.

In Mathematica,
In[1]:= Tan[Pi/2]
Out[1]= ComplexInfinity
that is, the infinity of C*, the onepoint compactification of C.
Quote:  It should give an error message since the limit does not exist.

It doesn't exist in the reals or their twopoint compactification. But it
does exist in their onepoint compactification and in C*, as Mathematica
correctly says.
Quote:  Left and right limits are +inf and inf.
I hope Maple doesn't give inf for limit{ x > pi/2 ; tan(x) } ?
Similarly with tanh(infinity) and arctanh(1). We must be
evervigilant, I suppose.

In[2]:= ArcTanh[1]
Out[2]= Infinity
which is also correct.
And speaking of _abus de langage_, arc<hyperbolic function> is a misnomer,
although often used nowadays.
David 

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Dirk Van de moortel science forum Guru
Joined: 01 May 2005
Posts: 3019

Posted: Sun Jun 25, 2006 8:47 pm Post subject:
Re: limit(ln(x+sqrt(x^2+1)),x=infinity)



"Gene Ward Smith" <genewardsmith@gmail.com> wrote in message news:1151266465.658389.154710@i40g2000cwc.googlegroups.com...
Quote: 
Dirk Van de moortel wrote:
I just don't think the concept of "extended reals" is useful in
real analysis.
My "attains to increasingly negative values etc" shows one use; that's
really too sloppy to be an actual definition. I can say "f(x) 
infinity" iff arctan(f(x)) > pi/2", but then I'm really using the
extended reals and may as well be explicit about it, projecting the
real numbers onto a semicircular arc and sticking in pi/2 and pi/2 as
the endpoints of the arc.
Fooling around with Maple, I see it takes values in the extended reals,
but won't give them. That is, arctan(infinity) = pi/2 according to
Maple, but tan(pi/2) returns an error message.

It should give an error message since the limit does not exist.
Left and right limits are +inf and inf.
I hope Maple doesn't give inf for limit{ x > pi/2 ; tan(x) } ?
Quote:  Similarly with tanh(infinity) and arctanh(1). We must be
evervigilant, I suppose.

yes, and I hear these packages have a few bugs as well :)
Dirk Vdm 

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Gene Ward Smith science forum Guru
Joined: 08 Jul 2005
Posts: 409

Posted: Sun Jun 25, 2006 8:14 pm Post subject:
Re: limit(ln(x+sqrt(x^2+1)),x=infinity)



Dirk Van de moortel wrote:
Quote:  I just don't think the concept of "extended reals" is useful in
real analysis.

My "attains to increasingly negative values etc" shows one use; that's
really too sloppy to be an actual definition. I can say "f(x) >
infinity" iff arctan(f(x)) > pi/2", but then I'm really using the
extended reals and may as well be explicit about it, projecting the
real numbers onto a semicircular arc and sticking in pi/2 and pi/2 as
the endpoints of the arc.
Fooling around with Maple, I see it takes values in the extended reals,
but won't give them. That is, arctan(infinity) = pi/2 according to
Maple, but tan(pi/2) returns an error message.
Similarly with tanh(infinity) and arctanh(1). We must be
evervigilant, I suppose. 

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Dirk Van de moortel science forum Guru
Joined: 01 May 2005
Posts: 3019

Posted: Sun Jun 25, 2006 8:08 pm Post subject:
Re: limit(ln(x+sqrt(x^2+1)),x=infinity)



"David W. Cantrell" <DWCantrell@sigmaxi.org> wrote in message news:20060625152159.846$LC@newsreader.com...
Quote:  "Dirk Van de moortel" <dirkvandemoortel@ThankSNOSperM.hotmail.com> wrote:
"David W. Cantrell" <DWCantrell@sigmaxi.org> wrote:
[snip]
But at least from my standpoint,
there is an _abus de langage_ if we say that "The limit is X." when X
is not an element of the system in which we are working.
Yes, I agree.
But it seems we have slightly different standpoints.
I just don't think the concept of "extended reals" is useful in
real analysis.
Authors of many texts in real analysis disagree with you; otherwise, they
wouldn't mention the extended reals. (Of course, there are also many
authors who don't.)
Looking at how the addition of these two
elements to the field of the reals creates havoc in the field
properties, quite on the contrary, I'd say.
For my standpoint, there is no havoc. The system of extended reals contains
the field of reals, with all of its properites unaltered.

I.m.o. the worst thing about them is the fact that addition and
multiplication stops being defined for all elements of the set,
which immediately requires making modifications to the
expressions of the field properties. For instance, some of the
occurences of "for all x..." would have to be replaced with
"for all x, except +inf and inf". Even worse, "for all x and y"
being replaced with "for all x and y, except when x = +inf
and y = inf, or x = inf and y = +inf..." etc...
And yes indeed, I have seen many authors mention them.
but then stop doing anything uselful with them after having
mentioned them :)
Dirk Vdm 

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Gene Ward Smith science forum Guru
Joined: 08 Jul 2005
Posts: 409

Posted: Sun Jun 25, 2006 7:59 pm Post subject:
Re: limit(ln(x+sqrt(x^2+1)),x=infinity)



David W. Cantrell wrote:
Quote:  AFAIK, no mathematicians would say that the limit _exists_ in such a case,
unless they were working in the extended reals.

If they claim the limit takes a value, and the value is infinity, it
seems to me they are in the extended reals whether they want to be or
not. If "the limit is infinity" is shorthand for
"attains to increasingly negative values with no lower bound", then
it's either a definition or an abuse of language. 

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David W. Cantrell science forum Guru
Joined: 02 May 2005
Posts: 352

Posted: Sun Jun 25, 2006 7:49 pm Post subject:
Re: limit(ln(x+sqrt(x^2+1)),x=infinity)



matt271829news@yahoo.co.uk wrote:
Quote:  David W. Cantrell wrote:
matt271829news@yahoo.co.uk wrote:
matt271829n...@yahoo.co.uk wrote:
Dirk Van de moortel wrote:
"Dirk Van de moortel"
dirkvandemoortel@ThankSNOSperM.hotmail.com> wrote in message
news:6ozng.502363$vj5.12673089@phobos.telenetops.be...
"David W. Cantrell" <DWCantrell@sigmaxi.org> wrote in message
news:20060625124832.349$2L@newsreader.com... "Dirk Van de
moortel"
dirkvandemoortel@ThankSNOSperM.hotmail.com> wrote: "Dirk
Van de moortel"
dirkvandemoortel@ThankSNOSperM.hotmail.com> wrote in
message
news:PQtng.501856$zH5.12480544@phobos.telenetops.be...
matt271829news@yahoo.co.uk> wrote in message
news:1151176925.925274.103960@p79g2000cwp.googlegroups.com.
..
N. Silver wrote:
Jiri Slaby wrote:
I have no clue, how to compute this limit without
maple: limit(ln(x+sqrt(x^2+1)), x=infinity)
Here's a heuristic approach.
As the magnitude of x gets large
sqrt(x^2+1) ~ sqrt(x^2) = abs(x).
All the while, sqrt(x^2+1) > abs(x).
Then ln(x+sqrt(x^2+1)) ~ ln(x+abs(x)).
As x goes to infinity, ln(x+sqrt(x^2+1))
approaches ln(0) from the positive side of
zero. So, the limit is minus infinity.
Slightly picky question: is it technically correct to say
"the limit is minus infinity"? I would be more inclined to
say that the limit doesn't exist at all...
Consider two contexts.
Context 1: We're dealing with just the reals
There is certainly no element "oo". If we say "The limit
is oo.", we are obviously not using the word "is" in an
ordinary way, as we are when we say, for example, "The
limit is 4." I would not think of "The limit is oo." as
actually being technically incorrect, but rather as using a
particular _abus de langage_ which is very familiar to us
mathematicians. If we had said only "The limit does not
exist.", we would have conveyed much less information.
Context 2: We're dealing with the extended real, [oo, +oo]
"The limit is oo." means exactly what it appears to mean.
The statement is technically correct. The limit exists.
There is no _abus de langage_.
Yes, of course it exists.
Dirk: I must suppose you were being sarcastic. Otherwise, you
surely
wouldn't have followed that comment with a definition
specifically designed to _avoid_ needing oo to exist.
No, I wasn't sarcastic at all.
As far as I'm concerned this has nothing to do with
"working in the reals or in the extended reals".
When the statement
 for each real M > 0, there is a real L > 0, such that
 for all real x in the domain of f, the following
 implication holds:
 x < L ==> f(x) < M
is true, we say that the limit exists and is infinity.
AFAIK, no mathematicians would say that the limit _exists_ in such a
case, unless they were working in the extended reals.
I can't imagine any mathematician I have ever met saying that
in that case "the limit does not exist unless one works in the
extended reals".
... and referring to your two contexts, I would add that
_abus de langage_ doesn't even have to come in the
picture either. This is actually the way we can *define*
language. The standard epsilondelta (and epsilonL
and Mdelta and ML) definitions allow us to work with
phrases like "infinite limits" and "limits for x going to
infinity" without ever even mentioning the "extended reals".
Your last sentence above is correct. But at least from my standpoint,
there is an _abus de langage_ if we say that "The limit is X." when X
is not an element of the system in which we are working.
There is no real need to look at +oo and oo as numbers
of any set.
Well, that is how we got this stuff at this side of the
Atlantic anyway :)
Dirk Vdm
If the phrase "the limit is infinity" is *defined* as you suggest
then I don't see there can be a problem.
Do you also think that there can be no problem with saying that the
limit actually _exists_ in such a case?
No, if we're talking about real numbers then I don't think that the
limit can be said to "exist"...

Good.
Quote:  which is where I came in. However, if
"the limit is infinity" is defined and generally understood to be
shorthand for what Dirk said then I don't see a problem.

The "problem", unless I severely misunderstood Dirk's position, was that he
said the limit _exists_ even if we're just talking about reals numbers.
Quote:  (Whether this
is actually the case I don't know.) I suppose you could argue that "is"
implies "exists",

One would normally think that to be the case!
Quote:  ergo contradiction, but that would be too extreme I feel...

Note that I didn't say anything that extreme. I just said it was an _abus
de langage_.
Quote:  AFAIK, regardless of the side of the
Atlantic you're on, such a limit does not exist (assuming that we're
not dealing with the extended reals).
We just have to accept that the
definition of "the limit is infinity" is different from the
definition of "the limit is x" for some real number x, right?
By this I mean that we can't just take the definition of "the limit
is x" and replace "x" by "infinity"...
Sure you can, if your definition of "the limit is x" was suitable.
I was referring to the usual definition when x is a real number.

If, in the usual definition to which you refer, we merely replace the two
instances of expressions in the form p  q, representing the distance
between p and q, by d(p, q), where d is a suitable metric, we get a
definition which works on the extended reals. We then have just _one_ limit
definition, rather than needing to have separate definitions to handle
special cases.
David 

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David W. Cantrell science forum Guru
Joined: 02 May 2005
Posts: 352

Posted: Sun Jun 25, 2006 7:28 pm Post subject:
Re: limit(ln(x+sqrt(x^2+1)),x=infinity)



David W. Cantrell <DWCantrell@sigmaxi.org> wrote:
Quote:  "Dirk Van de moortel" <dirkvandemoortel@ThankSNOSperM.hotmail.com
wrote:
"David W. Cantrell" <DWCantrell@sigmaxi.org> wrote:
[snip]
But at least from my standpoint,
there is an _abus de langage_ if we say that "The limit is X." when X
is not an element of the system in which we are working.
Yes, I agree.
But it seems we have slightly different standpoints.
I just don't think the concept of "extended reals" is useful in
real analysis.
Authors of many texts in real analysis disagree with you; otherwise, they
wouldn't mention the extended reals. (Of course, there are also many
authors who don't.)
Looking at how the addition of these two
elements to the field of the reals creates havoc in the field
properties, quite on the contrary, I'd say.
For my standpoint, there is no havoc. The system of extended reals
contains the field of reals, with all of its properites unaltered.

I meant to type "properties".
David 

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David W. Cantrell science forum Guru
Joined: 02 May 2005
Posts: 352

Posted: Sun Jun 25, 2006 7:21 pm Post subject:
Re: limit(ln(x+sqrt(x^2+1)),x=infinity)



"Dirk Van de moortel" <dirkvandemoortel@ThankSNOSperM.hotmail.com> wrote:
Quote:  "David W. Cantrell" <DWCantrell@sigmaxi.org> wrote:
[snip]
But at least from my standpoint,
there is an _abus de langage_ if we say that "The limit is X." when X
is not an element of the system in which we are working.
Yes, I agree.
But it seems we have slightly different standpoints.
I just don't think the concept of "extended reals" is useful in
real analysis.

Authors of many texts in real analysis disagree with you; otherwise, they
wouldn't mention the extended reals. (Of course, there are also many
authors who don't.)
Quote:  Looking at how the addition of these two
elements to the field of the reals creates havoc in the field
properties, quite on the contrary, I'd say.

For my standpoint, there is no havoc. The system of extended reals contains
the field of reals, with all of its properites unaltered.
David 

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Dirk Van de moortel science forum Guru
Joined: 01 May 2005
Posts: 3019

Posted: Sun Jun 25, 2006 7:06 pm Post subject:
Re: limit(ln(x+sqrt(x^2+1)),x=infinity)



"David W. Cantrell" <DWCantrell@sigmaxi.org> wrote in message news:20060625142627.600$nw@newsreader.com...
Quote:  matt271829news@yahoo.co.uk wrote:
matt271829n...@yahoo.co.uk wrote:
Dirk Van de moortel wrote:
"Dirk Van de moortel" <dirkvandemoortel@ThankSNOSperM.hotmail.com
wrote in message
news:6ozng.502363$vj5.12673089@phobos.telenetops.be...
"David W. Cantrell" <DWCantrell@sigmaxi.org> wrote in message
news:20060625124832.349$2L@newsreader.com... "Dirk Van de moortel"
dirkvandemoortel@ThankSNOSperM.hotmail.com> wrote: "Dirk Van
de moortel" <dirkvandemoortel@ThankSNOSperM.hotmail.com> wrote
in message
news:PQtng.501856$zH5.12480544@phobos.telenetops.be...
matt271829news@yahoo.co.uk> wrote in message
news:1151176925.925274.103960@p79g2000cwp.googlegroups.com...
N. Silver wrote:
Jiri Slaby wrote:
I have no clue, how to compute this limit without maple:
limit(ln(x+sqrt(x^2+1)), x=infinity)
Here's a heuristic approach.
As the magnitude of x gets large
sqrt(x^2+1) ~ sqrt(x^2) = abs(x).
All the while, sqrt(x^2+1) > abs(x).
Then ln(x+sqrt(x^2+1)) ~ ln(x+abs(x)).
As x goes to infinity, ln(x+sqrt(x^2+1))
approaches ln(0) from the positive side of
zero. So, the limit is minus infinity.
Slightly picky question: is it technically correct to say "the
limit is minus infinity"? I would be more inclined to say that
the limit doesn't exist at all...
Consider two contexts.
Context 1: We're dealing with just the reals
There is certainly no element "oo". If we say "The limit is
oo.", we are obviously not using the word "is" in an ordinary
way, as we are when we say, for example, "The limit is 4." I
would not think of "The limit is oo." as actually being
technically incorrect, but rather as using a particular _abus
de langage_ which is very familiar to us mathematicians. If we
had said only "The limit does not exist.", we would have
conveyed much less information.
Context 2: We're dealing with the extended real, [oo, +oo]
"The limit is oo." means exactly what it appears to mean. The
statement is technically correct. The limit exists. There is no
_abus de langage_.
Yes, of course it exists.
Dirk: I must suppose you were being sarcastic. Otherwise, you
surely
wouldn't have followed that comment with a definition
specifically designed to _avoid_ needing oo to exist.
No, I wasn't sarcastic at all.
As far as I'm concerned this has nothing to do with
"working in the reals or in the extended reals".
When the statement
 for each real M > 0, there is a real L > 0, such that
 for all real x in the domain of f, the following
 implication holds:
 x < L ==> f(x) < M
is true, we say that the limit exists and is infinity.
AFAIK, no mathematicians would say that the limit _exists_ in such a case,
unless they were working in the extended reals.

Then I think our mileages vary somewhat :)
Quote: 
I can't imagine any mathematician I have ever met saying that
in that case "the limit does not exist unless one works in the
extended reals".
... and referring to your two contexts, I would add that
_abus de langage_ doesn't even have to come in the
picture either. This is actually the way we can *define*
language. The standard epsilondelta (and epsilonL
and Mdelta and ML) definitions allow us to work with
phrases like "infinite limits" and "limits for x going to infinity"
without ever even mentioning the "extended reals".
Your last sentence above is correct. But at least from my standpoint, there
is an _abus de langage_ if we say that "The limit is X." when X is not an
element of the system in which we are working.

Yes, I agree.
But it seems we have slightly different standpoints.
I just don't think the concept of "extended reals" is useful in
real analysis. Looking at how the addition of these two
elements to the field of the reals creates havoc in the field
properties, quite on the contrary, I'd say.
This might be one of these Bourbaki things.
Dirk Vdm 

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matt271829news@yahoo.co. science forum Guru
Joined: 11 Sep 2005
Posts: 846

Posted: Sun Jun 25, 2006 7:04 pm Post subject:
Re: limit(ln(x+sqrt(x^2+1)),x=infinity)



David W. Cantrell wrote:
Quote:  matt271829news@yahoo.co.uk wrote:
matt271829n...@yahoo.co.uk wrote:
Dirk Van de moortel wrote:
"Dirk Van de moortel" <dirkvandemoortel@ThankSNOSperM.hotmail.com
wrote in message
news:6ozng.502363$vj5.12673089@phobos.telenetops.be...
"David W. Cantrell" <DWCantrell@sigmaxi.org> wrote in message
news:20060625124832.349$2L@newsreader.com... "Dirk Van de moortel"
dirkvandemoortel@ThankSNOSperM.hotmail.com> wrote: "Dirk Van
de moortel" <dirkvandemoortel@ThankSNOSperM.hotmail.com> wrote
in message
news:PQtng.501856$zH5.12480544@phobos.telenetops.be...
matt271829news@yahoo.co.uk> wrote in message
news:1151176925.925274.103960@p79g2000cwp.googlegroups.com...
N. Silver wrote:
Jiri Slaby wrote:
I have no clue, how to compute this limit without maple:
limit(ln(x+sqrt(x^2+1)), x=infinity)
Here's a heuristic approach.
As the magnitude of x gets large
sqrt(x^2+1) ~ sqrt(x^2) = abs(x).
All the while, sqrt(x^2+1) > abs(x).
Then ln(x+sqrt(x^2+1)) ~ ln(x+abs(x)).
As x goes to infinity, ln(x+sqrt(x^2+1))
approaches ln(0) from the positive side of
zero. So, the limit is minus infinity.
Slightly picky question: is it technically correct to say "the
limit is minus infinity"? I would be more inclined to say that
the limit doesn't exist at all...
Consider two contexts.
Context 1: We're dealing with just the reals
There is certainly no element "oo". If we say "The limit is
oo.", we are obviously not using the word "is" in an ordinary
way, as we are when we say, for example, "The limit is 4." I
would not think of "The limit is oo." as actually being
technically incorrect, but rather as using a particular _abus
de langage_ which is very familiar to us mathematicians. If we
had said only "The limit does not exist.", we would have
conveyed much less information.
Context 2: We're dealing with the extended real, [oo, +oo]
"The limit is oo." means exactly what it appears to mean. The
statement is technically correct. The limit exists. There is no
_abus de langage_.
Yes, of course it exists.
Dirk: I must suppose you were being sarcastic. Otherwise, you
surely
wouldn't have followed that comment with a definition
specifically designed to _avoid_ needing oo to exist.
No, I wasn't sarcastic at all.
As far as I'm concerned this has nothing to do with
"working in the reals or in the extended reals".
When the statement
 for each real M > 0, there is a real L > 0, such that
 for all real x in the domain of f, the following
 implication holds:
 x < L ==> f(x) < M
is true, we say that the limit exists and is infinity.
AFAIK, no mathematicians would say that the limit _exists_ in such a case,
unless they were working in the extended reals.
I can't imagine any mathematician I have ever met saying that
in that case "the limit does not exist unless one works in the
extended reals".
... and referring to your two contexts, I would add that
_abus de langage_ doesn't even have to come in the
picture either. This is actually the way we can *define*
language. The standard epsilondelta (and epsilonL
and Mdelta and ML) definitions allow us to work with
phrases like "infinite limits" and "limits for x going to infinity"
without ever even mentioning the "extended reals".
Your last sentence above is correct. But at least from my standpoint, there
is an _abus de langage_ if we say that "The limit is X." when X is not an
element of the system in which we are working.
There is no real need to look at +oo and oo as numbers
of any set.
Well, that is how we got this stuff at this side of the
Atlantic anyway :)
Dirk Vdm
If the phrase "the limit is infinity" is *defined* as you suggest then
I don't see there can be a problem.
Do you also think that there can be no problem with saying that the limit
actually _exists_ in such a case?

No, if we're talking about real numbers then I don't think that the
limit can be said to "exist"... which is where I came in. However, if
"the limit is infinity" is defined and generally understood to be
shorthand for what Dirk said then I don't see a problem. (Whether this
is actually the case I don't know.) I suppose you could argue that "is"
implies "exists", ergo contradiction, but that would be too extreme I
feel...
Quote:  AFAIK, regardless of the side of the
Atlantic you're on, such a limit does not exist (assuming that we're not
dealing with the extended reals).
We just have to accept that the
definition of "the limit is infinity" is different from the definition
of "the limit is x" for some real number x, right?
By this I mean that we can't just take the definition of "the limit is
x" and replace "x" by "infinity"...
Sure you can, if your definition of "the limit is x" was suitable.

I was referring to the usual definition when x is a real number.


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