Bob and Dick go working to earn money. At the end of the month Bob gets A $ and Dick gets 1.1*A $. What is the resolution of this paradox?

@Fausto

you should understand that extending (x,y) |--> x^y to the point (0,0) is not (primarily) a matter of considering one, two, or more functions but of selecting a curve (path) in R^2 which ends at (0,0). Of course, such a path again gives rise to a function

t |-->x(t)^y(t) and to the question whether this function has a limit at the path's end. As Robert showed, depending on the selected path, one gets infinitely many limits.

That not all paths give the same limit is one of the many ways to express the fact that the function z(x,y) has no continuous extension to (x,y)=(0,0).

(I have suppressed domains and codomains of the functions under consideration, since for functions resulting from arithmetic expressions there are natural choices for these.)

Your critique of this argument can only be understood as an outgrowth of wrath on downvotes. If there is incompetence in the game, then certainly not in Robert's contributions.

@Fausto,

in my view downvoting is a reasonable reaction to contributions which are trivial, false, or offending and formulated in a way that there is no much hope that the author will understand or follow an argument that points out this deficiency. The downvotes of your first contribution were certainly not justified by this criterion (none of your downvotes comes from me, btw). In my mind downvoting should be as public as upvoting.

We can see the problem as lack of equality between two limits of two different functions of a single variable. At least it is originally stated as such. But then - why paradox?

On the other hand we may see this problem as different limits of a single function of two variables when approaching (0,0) point. They are obviously different when approaching (0,0) from different directions. This only means the function has no true limit at (0,0), that's all. Otherwise the symbol 0^0 would have well defined numerical value, but this is not the case. When the function has different limits at a given point, depending on the approach trajectory, we call such a function discontinuous at this very point. But "normal" discontinuous functions have well defined values at discontinuity point (a "jump"). Here, however, the situation is worse: x^y simply doesn't exist at (0,0), or, in other words, (0,0) is not in domain of this function.

Is it still a paradox? Not for me, a physicists.

Oh Fausto,

Of course, the successive limit lim_{x-->0}(lim_{y-->0} x^y) is simply 1. But your words

"The limit for BOTH x--->0 AND y--->0 is ALWAYS=1" (which have no defined meaning) can hardly be interpreted to mean this. It is a blunder statement of the kind which even beginners of mathematics ought have learned to avoid. Since enough hints were given to you how things can be treated precisely it looks to me that you have to learn a lot before we can expect clear and correct mathematical arguments from you.

@Fausto,

indeed lim_{x-->0}(lim_{y-->0} x^y) is defined just as you interpreted it. It is a uniquely defined expression --- even if the word 'successive' would not have been used. I used it only as a help to those for whom the reading of mathematical terms works not as seamlessly as would be desirable. I don't think that you are stupid and ignorant, but I do think that you are ignorant of the formal structure of mathematical statements. Although this can be said of many active mathematicians, it is certainly worthwhile to strive for improvement.

That x tends to 0 isn't means x=0.

@@Fausto, note that the usage of all capitals words in messages is seen as screaming at someone's face. This is common net-etiquette nowadays.

If that were in person it would be offensive and certainly far from instructive or helping, don't you think? My humble recommendation is to use them sparingly.

For emphasis, I prefer to use suggestive _italics_, although the mere emphasis can also be treated cautiously.

Again, as @Ulrich pointed out, the whole point is to provide help.

Don't give up @Robert.

@Fausto, please consider the following succession of points in the plane

x_1=e^{-10}, y_1=7/10

x_2=e^{-100}, y_2=7/100

x_3=e^{-1000}, y_3=7/1000

...

x_n=e^{-10^n}, y_n=7/(10^n)

...

and now be so kind to operate (x_n)^(y_n) to obtain (e^{-10^n})^{7/10^n}=e^{-7}, which is not 1.

Since Fausto seems not to hear what people say, I hope he will be inspired by what machines say. Here is the voice of Mathematica.

http://www.ulrichmutze.de/pedagogic_stuff/fausto.pdf

@Fausto

In the "proof" in your document you seem to miss that the curved triangular open set F_\epsilon is not a neighborhood of the point (0,0). This property, however, would be needed to prove the correctness of your "definition" of the limit.

The Mathematica based demonstration in the link treats a similar case to that under discussion here. That it came out just this week seems to be mere coincidence.

http://www.demonstrations.wolfram.com/NonexistenceOfAMultivariableLimit/

Oh Fausto, all your contributions to this discussion proved your incompetence in the rather elementary field of mathematics we are discussing. So you should better not tell us that you ever spoke about quality and about incompetent professors. Sounds ridiculous, really.

The definition of limit (of a function, of a succession) is well stablished and upon this ground we give logic reasoning to obtain our results, which are in our case that 1 is not the limit of x^y at (0,0) (in fact, that x^y has no limit at (0,0)). This result stands by itself and is not related with opinions, qualities or something stablished by some particular one acting like Humpty Dumpty(*), dear Fausto. Six days ago you fall in mistake. It was explained to you by some colleagues with many logical arguments. Now you introduce a new concept (something you call quality) like a decoy. If you are a real scientist you ought to recognize the truth.

(*)"When I use a word,' Humpty Dumpty said in rather a scornful tone, 'it means just what I choose it to mean"

@Fausto

No, I did'nt. My judgement is based solely on what you gave us to read here.

@Fausto Galetto:

My critique was about the set F_\epsilon not being a neighborhood of (0,0). This is a statement about sets and not about functions. So I do not understand your question, as in my comment no functions were involved.

Anyway, what I do not understand is what you are fighting for with such an intensity. Do you want to solve a "paradox" for which nobody cares - since it is not a paradox in the first place? See, if you want to provide a reason for saying that 0^0 = 1, there is no need to do so. In certain areas of mathematics it is perfectly valid to say that 0^0 = 1. In other areas of mathematics that is a very bad idea, and there are several good reasons for that.

Modern mathematics works _only_ by the following principle: Starting from a set of axioms (statements that are _said to be true_) by purely deductive logical reasoning (and the rules for that are very strict) new statements are proved. So, if you want to do mathematics you have to follow that path. Otherwise, you are doing something else, i.e. not mathematics.

Depending on the set of axioms you start with you get a formal system, in which certain other statements can be proved or disproved - if your formal system contains enough power to work with the natural numbers there is always also the class of statements that can be neither proved nor disproved.

You can choose your set of axioms for your formal system in principle as you like _but_ you should make sure that your axioms do not contain contradictions, as otherwise _all_ statements in your formal system can be proved (and disproved).

A definition is just a way to assign a name to a given formal object. You do not get anything true or false by a definition. You can give useful or useless names, however.

Back to the x^y discussion:

If you talk about real valued functions depending on real arguments with mathematicians, it will be implicitly understood that the formal system you work with are (at the very least) the axioms of the field of real numbers. If you then write down symbols like lim (meaning limit) it is also implicitly assumed that you conform with the definition of lim that is used by all mathematicians. (Note that this is useful. It is the same principle that is used when you build or drive cars. While it is from a technical point perfectly possible to use the left and right pedals in your car for steering to the left and to the right, respectively, and the (steering)wheel in front of you for acceleration when turning to the left and for breaking when turning to the right, no car manufacturer would implement this design, and I guess you can understand perfectly well why. So agreeing on common basic principles is important.) Thus, if you talk about mathematics, we all will (actually _must_) assume that you are using the same logic, the same axioms, and the same definitions as we all do.

But then the following statements are true:

A) lim_{(x,y)->(0,0)} x^y = 1 if and only if

for _every_ sequence (x_n,y_n) with lim_{n->\infinity} (x_n,y_n) = (0,0) we have that lim_{n->\infinity}{x_n}^{y_n} = 1.

B) For the sequence (x_n,y_n)=(e^{-10^n},7/(10^n)) we get lim_{n->\infinity}(e^{-10^n},7/(10^n))=(0,0) and lim_{n->\infinity} (e^{-10^n})^(7/(10^n))=e^(-7).

C) e^(-7) not equal 1.

Now elementary logic tells that the second part of A (after the if and only if) is false, since not for every sequences with lim_{n->\infinity} (x_n,y_n) = (0,0) the limit lim_{n->\infinity}{x_n}^{y_n} is 1, because we have found a sequences in B for which the limit (by C) is different. Since A is an equivalence, by elementary logic this has the consequence that the first part of A (before the if and only if) is false. Therefore, we have proved that in the formal system of the real numbers the limit lim_{(x,y)->(0,0)} x^y is not 1. Actually, it can be reasoned in a very analogous way that for every real number a the limit lim_{(x,y)->(0,0)} x^y is not equal to a. Thus, the limit does not exist.

This, however, is not a paradox, since there is no contradiction to itself involved (see definition of a "paradox"), and it is not even a problem! There are (in a mathematically precise sense) many more functions which do _not_ possess a limit for (x,y)->(0,0) than there are functions which do possess such a limit, so why fight for a single one, for which the proof of non-existence of the limit is so easy?

Well, if you want to insist (in _your_ formal system) on the truth of the statement lim_{(x,y)->(0,0)} x^y = 1 you have basically two choices:

1. You can change the definition of the symbol lim - but this is comparable to use pedal steering and wheel acceleration in a car, and nobody apart from you will use it.

2. You can set the statement lim_{(x,y)->(0,0)} x^y = 1 formally to _true_ (i.e. make it an axiom) but then you get a contradiction within the axioms of your formal system, and suddenly all statements will become true and false at the same time, and again nobody apart from you will use the system.

_But_ if you only want to have that 0^0 = 1 there is no problem with that. You can stick to the natural numbers. Usually 0^0 = 1 is defined there, since in many cases (recursion, etc.) this is very useful, and there is no problem with limits in N.

You can even set 0^0 = 1 in the real numbers if you want. Then the function x^y will be defined at the point (0,0), and it will be discontiuous there, also not a problem.

Therefore, my advise to you is: Find a real interesting problem in the field of mathematics, e.g. Goldbach's conjecture. If you manage to come up with a proof of that one (vaild logical reasoning, without redefinition of "prime number" or "+", etc.), you will definitely become very famous, and, even if you do not manage to find a proof, eventually you will (if you are working diligently and arrive at the (valid!) conclusion that not all other mathematicians in the world are wrong and/or just rejecting you for the sake of i-don't-know-what) learn an awful lot about mathematics on the way - and you could study Lao Tse who might tell you, why this is not a completely bad idea in the first place.

@Fausto Galetto:

I am aware that you did not ask the question. However, you have made the impression that you want to solve it.

As I have explained, there is no need to solve the question, since it is already solved, as Robert J. Low had pointed out in his very first answer in the discussion.

I am aware that you have been downvoted several times. Although, I am not a friend of downvoting, I could explain to you probable reasons: First, you are repeating statements that are not compatible with mathematical facts. Second, you are constantly violating the netiquette by SHOUTING at people - although you have been told that this is considered impolite. I am quite sure that the downvoting will continue unless you change that behaviour.

Mathematicians are (in my experience taken from many conferences) mostly benevolent scientists - by the way in considerable contrast to physisists and biologists where I have witnessed really aggressive behaviour during discussions. One reason for that is that _usually_ it can be decided without doubt whether a statement that was made is correct or wrong. There is no room for "opinion", and so there is no big reason for quarrels. If someone is not sure about a statement, he can always ask his discussion partner to prove it, to give more details, etc., until the last bit of doubt is gone. At least that is the ideal situation (but we mathematicians love to idealize things...).

There are, however, a few things that cause the most patient mathematician to lose his temper. One of them is, if someone does not want to accept (and adopt) the way modern mathematics is performed (with proofs, agreed upon definitions, correct logic, correct terminology, using correct mathematical expressions - and the mathematical language is very strict), and if he continues to do so, no matter what is explained to him.

My advice is not to take the downvoting as an offence but to take it as a hint that there is room for improvement in your way of writing, in your way of formulating mathematical arguments, and in your ability to accept that you have made a mistake.

@Fausto Galetto:

That a point x is a cluster point of a set A is not enough to conclude that the set A is a neighborhood of x. Even if the set is simply connected and closed that is in general wrong, even in a situation where x=0 and the space is the first orthant. The point is that in two or more dimensions you have to make sure that your set A covers all "directions that can be used to approach x". It must not be possible to approach x from outside A by a curve that completely avoids A (except for meeting A at x). This situation can never happen in R^1 but in R^2 it can happen. E.g., the set

T={(x,y) | x>=0, y>=0, x=0} to approach (0,0) without ever meeting T until x is reached, i.e., g and T intersect only in (0,0).

For a set A to be a neighborhood of x you must make sure that there exists an \epsilon>0 such that _for all_ y (here in the first orthant) with ||x-y||

@Robert J. Low

Your example is a really nice one.

However, I was talking about sets and neighborhoods since Fausto Galetto used this argument about his F_\epsilon being similar to (0,1) or something like that, and I was just trying to make the point that sets in 2D near corners like (0,0) in the first quadrant are geometrically significantly different than 1D sets near 0 in the nonnegative real line, and I think that getting a similar behaviour as in your example with simply connected sets is not possible in R^1.

@Fausto Galetto

Two things after I read the Free Gazelle's last will: A neighborhood U of x is a set that contains the point x in its open _interior_.

You say that "As a matter of fact, letting G={0

While I accept that computer calculations use algorithms that may not exactly represent the problem, and having noted Dr Carriegos' example, if you create, say, 50 values x=h*random(0,1), y=h*random(0,1), calculate x^y for each, then the Minimum and Maximum of the results both converge towards 1 as h is reduced. This happens also if the 'h's are very different.

It is well-known that Monte-Carlo methods often significantly underestimate worst-case situations (in this case the curves tending to 0). A MC method is very good in detecting the most probable situation but usually not so good in detecting extremal situations.

There are so many functions f(x,y) which are not continuous in (0,0). There is nothing special about this one. Take for example f(x,y)=x/y. You can have any answer depending on the way you let x and y go to zero. It is trivial that f -> a along lines x=ay.

I have greatly enjoyed reading these posts! Yet again am I reminded of just how slippery the limit concept can be, and how I struggled with it. And still do...

@Fausto:

I am using definitions (and an easy analogon) from your paper "Quality Engineering vs “mathematicians”-QFG case n° TWENTY ...":

For f:\R^2->\R:

You say (in principle) that the limit lim_{x->u} f(x) = L if chosen a number \epsilon>0 there is an open disk D around u with radius \delta(\epsilon)>0 such that for all y\in D one has |f(y)-L|\R^2 we say that the limit lim_{x->u} g(x) = M if chosen a number \epsilon>0 there is an open interval D around u with radius \delta(\epsilon)>0 such that for all y\in D one has ||g(x)-M||u} f(x) = L and lim_{x->v} g(x) = u, then lim_{x->v} f(g(x)) = L.

Proof:

Let \epsilon>0 be arbitrary, as small as you like. Then, because lim_{x->u} f(x) = L, we get an open disk D_1 around u with radius \delta_1(\epsilon)>0 such that for all z\in D_1 one has |f(z)-L|v} g(x) = u, we get an open interval D around v with radius \delta(\delta_1(\epsilon)) such that for all y\in D one has ||g(y)-u||0} g(x) = (0,0):

Take \epsilon>0 arbitrary. Then we use the interval [-\delta,\delta] with radius \delta:=min(\epsilon / \sqrt{2},e^{-2 / \epsilon^2}) and see that for |x|0} e^{ln{|x|} / \sqrt{|ln(|x|)|}}

= 0

(for the last equality use for 1>\epsilon>0 the interval [-\delta,\delta] with \delta=e^{-log(\epsilon)^2}).

So with your definitions you have managed to introduce the equality 0=1 which has the trivial consequence that all numbers are equal.

Very well!

Quality engineering has proved its merits and has truly won against the "mathematicians". And citing your last publication again: Many wrongs do not make a right!

And adding a cite of my own: Half-knowledge is dangerous! You could include that into your future publications...

What was your vicious circle again?

Presumptuousness: Going beyond what is right or proper because of an excess of self-confidence or arrogance.

Ignorance: The condition of being uninformed or uneducated. Lack of knowledge or information.

Self-assessment is the first step for improvement!

No more comments from my side.

@Fausto Galetto,

A. It does not matter, whether I have read the pages 14 to 21, since, no matter what you write there, the fact remains true that your insistence of the limit being 1 implies that 0=1 - and slowly I begin to wonder, whether I am on a strange online version of Candid Camera, because nobody can be so bullheaded that he does not understand simple logic.

B. Most of the things you write on pages 14 to 21 are a mixture of trivialities and nonsense, e.g. that rot(grad(f)) = 0 is true in general if the function is C^2, and you prove it every time. That closed curve integrals in gradient fields are 0 is also true in general, and for proving that you do not need the rot(grad(f))=0 property nor the Stokes theorem, since it is a simple consequence of the gradient theorem. Starting from there everything is nonsense: You basically use hand waving to draw a connection between the computation of a limit and the computation of an equipotential line - and you impose the constant of the equipotential line in such a way that the result you wants falls out. This is especially eye-popping in the case where your field is x^y (why do you not consider the equipotential line for, e.g., k=0.1 in this case?). In addition, you claim that the potential y ln(x) is 0 at (0,0) which is equally wrong - you just make the same mistake as the one with x^y and its limit again.

C. This is really my final comment. By now I have seen more of the discussions you take part in (e.g. the Bayesian challenge), and I finally realize what most of my colleagues apparently have already realized: It is futile to discuss with you, because you deliberately suppress facts that do not fit into your view of the world, and you are either unable or unwilling to learn from your mistakes and/or admit that you erred. As a consequence: Quality must be loved, DISQuality must be ignored!

@Fausto Galetto,

The amount of ignorance is unbelievable. Please think and learn BEFORE you type!

I am still trying to explain things to you and hope that you do not continue to refuse to think.

You need to show rot v = 0 for a given vector field v on a simply connected region in order to prove that there exists a potential f with v = grad f. So you START with a vector field v and WANT TO KNOW whether it is a gradient field.

But you have already started with a gradient field, which is always irrotational, and a potential exists because YOU STARTED with a potential IN THE FIRST PLACE!

Nevertheless, you constantly IGNORE the most important argument, namely that the fact that you want the limit to be 1 logically implies that 0=1! But you don't seem to be able to grasp this fact.

Herrman,

what you describe as 'unbelievable', is according to my experience a wide-spread phenomenon. Using mathematics in a way that it helps to solve problems (instead of multiplying them) seems to be harder an art than those gifted individuals who never had major problems with it tend to acknowledge. Inspecting more recent mathematical textbooks and pedagogical math material in the web very probably creates the impression that the principle "garbage in - garbage out" is more often in action than not. Unfortunately there are many professions in science, engineering, quality management, and administration where mathematical sciolism is the source of real danger.

Dr. Abbott likes easy games and from time to time he presents one. I happened to read this game accidentally. Under a deceptive layout he has considered two different functions: 0^x with x that tends to 0 and x^0 with x that tends to 0.

The two functions are different and it is normal that the two limits are different. I see no paradox.

No problem. You are right and downvoters are wrong.

Ulrich,

I can understand that using mathematics and knowing what goes on is a difficult art, and I like to help out in these cases and I try to explain stuff, and I try to adapt my explanations to the knowledge of the other.

I completely agree with your remark on mathematical scholism.

What I find unbelievable here is the stubborn refusal to accept a mistake (which is not a shame - everybody errs, and it is important to learn from one's mistakes), the impolite and offensive style of discussion, and the mix-up of critical thinking (which is important) and denial of the truth.

I agree with Robert J. Low. In my opinion, if Limit_{(x,y)->(0,0)}f(x,y)=A exists, then whatever  the way and direction of (x,y) tends to (0,0) is, A is unique. If take x=0 and let y->0, or take y=0 and let x->0, they are only two direction, along x-axis or y-axis,  If along the two direction A is not unique. then the limit is not exist.

the limit does not exist

This thread consists of some of the dullest sciolism in the history of prose, a standardized academic jargon and rhetoric, the dutiful rehearsal of received theory, and the deliberate misrepresentation of anything challenging or rejecting academic postmodernism.

@Albert

> really impressive!

@Albert

I thought that even deconstructivists and other postmodern philosophers occasionally use footnotes and citations to flesh out their texts. Without those the remark is virtually useless.

After a minority reading I will present now a majority solution (I didn't read all preceding comments and I don't know if somebody has already presented this answer).

Anyway the function with two variables  z=x^y  has in the point with coordinates (0,0) an indetermination  0^0.  If we assume the variable y=0  and let us do tend the variable x to zero, the function  z=(limx^0) assumes the value 1, if instead we assume the variable x=0 and let us to tend the variable y to zero, the function  z=0^(limy)  assumes the value 0.

The existence of the two different limits is due therefore to the presence of an indetermination in the point of coordinates (0,0).

@Daniele Sasso:

Saying that there is "the presence of an indetermination" mathematically means that the limit does not exist (I think at least, since you did not properly define the term "indetermination" - there is something that is called indeterminate form or expression, and 0^0 is in the context of limits an indeterminate form. However, the presence of an indeterminate form does not mean that a limit does not exist. It only means that one needs to be very careful at that point).

One important fact: In mathematics we do not have minority and majority solutions. We have true and false statements and some indecidable statements.

Democracy is not a basic principle of mathematics. There are no compromises.

The next time when a bridge or a hospital is being built, we will perform a voting to decide whether it will collapse or not, should we?

In functions with one only variable f(x), when there is a discontinuity (indetermination) of first type in xo , for example functions with step, there isn't one limit in xo but there are two different limits: the right limit and the left limit. The same can happens for functions with two variables in points (x,y) in which there are discontinuities or indeterminations of first type. The function sinx/x has an indeterminacy in x=0, but in that case the two limits, right and left, match. Mathematics is beautiful because it is varied.

For the mathematical easy game presented by Abbott it would be also interesting to see what happens in points (k,y) and (x,k) where k is any constant number and doing tend x and y to zero in the two cases.

I agree that in mathematics and in science, and in generally in all intellectual activities, a principle of democracy is not valid in order to assess the cultural value of the work but its social value depends strongly on the judgement of the community. On the other hand also projects presented by engineers and architects must be in advance approved by community and if a doctor of medicine committs errors he is going towards heavy consequences.

If  0 1, then lim (a^{-n}) = 0 and lim (-1/n) =0, but

lim (a^{-n})^{-1/n} = a.

Now lim 1/n = 0, but  lim (1/n)^{1/n} = 1. 

Therefore, lim ((x,y) -> (0,0)) x^y don't exists. 

Dear All, the argument of Robert Low, I think, is not true.

For such a case it must hold that the line towards (0,0) must pass from (0,0) and not just to tend there. So, since y(x)=-k/ln(x) does not pass from (0,0), the argument is not applicable here.

Take for example the limit of (x/(x^2+y^2)*exp(-(x/(x^2+y^2))^2) as (x,y)->(0,0).

Then we can move at the line:

x/(x^2+y^2)=1 => x^2+y^2 = x => (x-1/2)^2+y^2=(1/2)^2,

which is a circle of center (1/2,0) and radius 1/2 and clearly passes from (0,0), so we can apply such an argument and prove that the example - limit does not exist.(See attachment for details)

But for the original limit this approach cannot be applied.

Demetris,

1. The line proposed by Robert Low indeed doesn't contain (0,0) point, but comes to it arbitrarily close. The lack of this single point (why not to add it to the original curve?) is immaterial when we discuss limits.  Therefore Robert's argument seems correct to me.

2. Your example is so nice that I took the liberty to enrich it with a quick and dirty sketch, see attachment.  Your circle is almost visible.  For clarity I didn't extend x-axis in negative direction too much.  If I did, you would be able to see another circle, located at  (-1/2,0) and with radius equal to 1/2.

Last but not least: f(circle)=const,   but   const=sign(x)*exp(-1), not 1.

I just wonder why the answer to the above problem is not the next and simple one:

*Let's go to (0,0) along the y-axis, then we have that f(0,y)=0^y=0-->0

*Let's go to (0,0) along the x-axis, then we have that f(x,0)=x^0=1-->1 # 0, so the limit does not exist.

Why do we have to impose such a y=-k/ln(x) and a new limit?

BTW, if we take f(x,y) in the domain [theta,oo) x [theta,oo) then we have that it tends to 1, see attachment for the case of theta=0.001.

(if we define f at that domain, then its minimum value is always theta^theta which can be easy proved that tends to 1 when theta --> 0)

@Demetris

A plot is not a proof, as you can easily see, because the plot is not correct along the y axis. The problem with the plot is that the computer interpolates essentially the results along points that are constructed parallel to the axes.Along those lines you cannot see the real behaviour of x^y close to the point (0,0).

Please read the various answers for an explanation, why the limit does not exist. It is basically connected to the definition of the limit. For a limit to exist - by definition - the result lim_n->oo f(x_n,y_n) must be 1 for all sequences (x_n,y_n) that converge to (0,0). There are sequences for which the limit does not converge to 1 as has been demonstrated by various colleagues in this forum, so the limit cannot be 1. (There are really good reasons why the limit is defined in such a way, but only a good book on analysis has enough space to explain that).

In addition, if you would insist on the limit to be 1 you could prove from that definition that 0=1 which would be a really bad contradiction, because it would have as a trivial consequence that all numbers are equal.

@Fausto Galetto

you were downvoted - I am sure - because mainly of how you say things, and because you apparently insist on replacing mathematics by some other thing - lets call it Galettomatics. This construct of yours does not use standard logic (because you just ignore necessary conditions for the theorems you use just because they do not fit your theories), it does not use the standard definitions (because otherwise it would be obvious that things do not work out as you wish them to be), and you just ignore counterexamples to and arguments against your theory because they are inconvenient for you. And your system contains contradictions which do not seem to trouble you.

And by the way in your last publication (in mathematical terms and not in Galettomatical terms) - with the Laplace transform - you make the same mistake again, and you could have had it much easier, because -ln s^y = ln (1/s)^y - setting x=1/s you get ln x^y, and you are back from where you started. Bad thing is: the limit lim_(s,y)->(\infinity,0) -ln s^y does not exist, and the reasons are basically the same as for x^y.

@Hermann, the plot refers for the specific domain [theta,oo) x [theta,oo) with theta very small, so the minimum value of f(x,y) is always theta^theta, or t^t, which tends to zero as t-->0:

tt=et ln(t)

so we search for the limit of t ln(t):

t ln(t) = ln(t) / (1/t) so by using L' Hospital rule the limit is the same with the limit:

(1/t) / (-1/t^2)=-t --> 0 and et ln(t) --> e0 = 1

On the other hand the limit clearly does not exist, neither for (0,oo)x(0,oo) nor for [0,oo)X[0,oo) because we can find two ways going to (0,0) with a different value for the limit.

So for the general case lim f(x,y) = [0,1], takes all the values between 0 and 1, analogously to the way you reach (0,0).

I think that the cause of conflict was that there exists a different view between an engineer like Fausto, who will set a minimum value theta for both x,y and a mathematician, who will examine the whole picture.

Anyway:

• the limit does not exist
• if we define f at the [theta,oo) x [theta,oo) with theta very small, then the 'limit' is 1
• for the general case the limit can take all values between 0 and 1

Dear Marek, thank you for your nice plot and extension to my presented example. I hadn't thought that I could also use the left circle. You see, I was happy enough to find one way, but your remark tells us the whole story.

Thanks!

@Fausto Galetto

Please, before you send answers check the books first!

lim_z->\infinity f(z) = lim_z->0 f(1/z) in the complex numbers, and if the limit exists in the complex numbers it must also exist in the real numbers (they are a subset of the complex numbers!). Since the limit does not even exist in the real numbers, it cannot exist in the complex numbers as well.

What you obviously do not understand is that the existence of the limit does not have anything to do with "WHY all you do not accept that y=-k/lnx MEANS that you impose the limit to stay on the plane z^(-k)? Do you like to do that? I do not.............".

Why do you not accept that this is just the second easiest counterexample to your claim, as you do not accept the trivial one of the original question. In reality there are infinitely many sequences that do not "stay on a plane" and where x_n^y_n does not converge to 1.

Use, e.g., x_n=1/n, y_n=1/ \sqrt{|ln(1/n)|}.

@Fausto Galetto

... unless z=0. This is so, because the complex numbers are a field. So what? How is this remark connected to my arguments? Do you assume that a limit that does not exist in the real numbers suddenly exists in the complex numbers?

Did I ever call you stupid? What I said is that - for some reason I do not understand - you seem to believe that everything that you say is correct, no matter how often other people prove to you that you are wrong.

I am only rather offensive, because you started it. Your whole publication is rather offensive. What goes around, comes around.

What you should appreciate is that I am still trying to explain to you what your misconceptions are, and that I am still talking to you, in contrast to most other people in this forum. And I am getting a bit more verbose occasionally because I want to wake you up and try to get you to reading and understanding the mathematical theories properly.