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Robert B. Israel
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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
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 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 ever-vigilant, 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
Robert B. Israel
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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)

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 ever-vigilant, 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
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 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.
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" 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.
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@ThankS-NO-SperM.hotmail.com> wrote:
 Quote: "Gene Ward Smith" 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 one-point 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 two-point compactification. But it
does exist in their one-point 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 ever-vigilant, 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,

David
Dirk Van de moortel
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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 ever-vigilant, I suppose.

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

Dirk Vdm
Gene Ward Smith
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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
ever-vigilant, I suppose.
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" wrote: "David W. Cantrell" 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
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.
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)

 Quote: David W. Cantrell wrote: matt271829-news@yahoo.co.uk wrote: matt271829-n...@yahoo.co.uk wrote: Dirk Van de moortel wrote: "Dirk Van de moortel" dirkvandemoortel@ThankS-NO-SperM.hotmail.com> wrote in message news:6ozng.502363\$vj5.12673089@phobos.telenet-ops.be... "David W. Cantrell" wrote in message news:20060625124832.349\$2L@newsreader.com... "Dirk Van de moortel" dirkvandemoortel@ThankS-NO-SperM.hotmail.com> wrote: "Dirk Van de moortel" dirkvandemoortel@ThankS-NO-SperM.hotmail.com> wrote in message news:PQtng.501856\$zH5.12480544@phobos.telenet-ops.be... matt271829-news@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 epsilon-delta (and epsilon-L and M-delta and M-L) 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
David W. Cantrell
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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" 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
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@ThankS-NO-SperM.hotmail.com> wrote:
 Quote: "David W. Cantrell" 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
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: matt271829-news@yahoo.co.uk wrote: matt271829-n...@yahoo.co.uk wrote: Dirk Van de moortel wrote: "Dirk Van de moortel" wrote in message news:20060625124832.349\$2L@newsreader.com... "Dirk Van de moortel" dirkvandemoortel@ThankS-NO-SperM.hotmail.com> wrote: "Dirk Van de moortel" wrote in message news:PQtng.501856\$zH5.12480544@phobos.telenet-ops.be... matt271829-news@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 epsilon-delta (and epsilon-L and M-delta and M-L) 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
matt271829-news@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: matt271829-news@yahoo.co.uk wrote: matt271829-n...@yahoo.co.uk wrote: Dirk Van de moortel wrote: "Dirk Van de moortel" wrote in message news:20060625124832.349\$2L@newsreader.com... "Dirk Van de moortel" dirkvandemoortel@ThankS-NO-SperM.hotmail.com> wrote: "Dirk Van de moortel" wrote in message news:PQtng.501856\$zH5.12480544@phobos.telenet-ops.be... matt271829-news@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 epsilon-delta (and epsilon-L and M-delta and M-L) 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.

 Quote: David

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