Can we use removable discontinuities to extend a function to the entire plane?

[FallenApple]

So we know that we typically have to use epsilon delta proofs for determining a limit of a multivariable function because there are infinite paths. But can we use removable discontinuities to prove a limit?

Say we want to evaluate the lim( x^2-y^2)/(x+y) as (x,y)->(0,0).

we can factor as follows (x-y)(x+y)/(x+y)

the x+y cancel and we are left with x-y

lim (x-y)=0 as (x,y)->(0,0) since x-y is clearly continuous a continuous function along any path.

Does this work?

What about the path y=-x? Is that entire line a removable discontinuity?

 

[jambaugh]

You raise an interesting question and I would say that the answer depends on the exact form of the definition. If the two variable limit is defined as an iterated limit,
$$\lim_{(x,y)\to(0,0)} f(x,y) \equiv \lim_{x\to 0} \left[ \lim_{y\to 0} f(x,y) \right]$$
then your example would have a limit since the line of discontinuity is not parallel to the coordinate axes.

If you use a full blown $\delta, \epsilon$ definition in terms of the positions as vectors then the limit would not be defined since the function is not defined for all
$$\mathbf{r} = \langle x,y\rangle: 0 < \lVert \mathbf{r}-\mathbf{r}_0\rVert < \delta$$
so you cannot have all the values are within epsilon of the limit for sufficently small delta for every position.

Consider if you adapted the definition to allow your example to have a limit, you would be effectively allowing that "for all $(x,y)$ within $\delta$" that the function is either within $\epsilon$ of the limit or undefined and that opens up some worse pathological examples. For example a function that is not defined within say an entire open region would be, by this modified definition, defined to have a limit equal to any value you choose (at the same time!) for a point in its interior since as you approach you're withing an arbitrarily small distance of the limit or the function is not defined, since the function is not defined for the entire neighborhood.

So I don't see how one can fix the definition to allow for a limit on such a point.

 

[lurflurf]

^Limits do not care what happens at the limit point.

if
g is continuous at x=a
and
for all x except x=a
f(x)=g(x)
then
$$\lim_{x\rightarrow a}\mathrm{f}(x)=\mathrm{g}(a)$$

we do not care about the value f(a) or if it exists

edited to add
The key words are removable discontinuities.
The cases you have in mind are non-removable discontinuities.
A frequent error in calculus is show a limit has value L if it exists and then conclude it has value L.
By definition if the we have a removable discontinuity the limit exist so we cannot have made that error. Though we may have made errors in finding L or concluding the discontinuity is removable.

 

[lurflurf]

What about the path y=-x? Is that entire line a removable discontinuity?

No you can only remove the point.
It is true that we use removable discontinuities to prove a limit.
You example is not a case that statement applies to since
$\frac{x^2-y^2}{x+y}=x-y$
fails in every neighborhood of zero.
it would need to be true in at least one.

 

[FallenApple]

No you can only remove the point.
It is true that we use removable discontinuities to prove a limit.
You example is not a case that statement applies to since
$\frac{x^2-y^2}{x+y}=x-y$
fails in every neighborhood of zero.
it would need to be true in at least one.

That was my hunch. Along that line itself, we can't approach that point of (0,0)

 

[FallenApple]

You raise an interesting question and I would say that the answer depends on the exact form of the definition. If the two variable limit is defined as an iterated limit,
$$\lim_{(x,y)\to(0,0)} f(x,y) \equiv \lim_{x\to 0} \left[ \lim_{y\to 0} f(x,y) \right]$$
then your example would have a limit since the line of discontinuity is not parallel to the coordinate axes.

If you use a full blown $\delta, \epsilon$ definition in terms of the positions as vectors then the limit would not be defined since the function is not defined for all
$$\mathbf{r} = \langle x,y\rangle: 0 < \lVert \mathbf{r}-\mathbf{r}_0\rVert < \delta$$
so you cannot have all the values are within epsilon of the limit for sufficently small delta for every position.

Consider if you adapted the definition to allow your example to have a limit, you would be effectively allowing that "for all $(x,y)$ within $\delta$" that the function is either within $\epsilon$ of the limit or undefined and that opens up some worse pathological examples. For example a function that is not defined within say an entire open region would be, by this modified definition, defined to have a limit equal to any value you choose (at the same time!) for a point in its interior since as you approach you're withing an arbitrarily small distance of the limit or the function is not defined, since the function is not defined for the entire neighborhood.

So I don't see how one can fix the definition to allow for a limit on such a point.

Makes sense. In two dimensions, we want to be able to approach the limit from all directions in the 2d plane, regardless of where.

 

[mathwonk]

with all due respect i disagree. I maintain that the limit does exist and has the suggested value, at least with the usually accepted definition of limit as found in what I consider standard books. the confusion is that for a limit to exist as (x,y)-->(0,0), one takes for granted that the values (x,y) only approach (0,0) along points where the function is defined. So even though the function is not defined on a line through (0,0) this has no effect on the value of the limit. i.e. for all epsilon, there is a delta such that for all (x,y) which are both within delta of (0,0) and lie in the domain of definition of the function, the value of the function at (x,y) is within epsilon of zero.

the only restriction usually made on this definition is that the point (0,0) should be a limit point of the domain. Here it is so. I.e. there is no problem with the function being undefined somewhere on every neighborhood of (0,0), rather the key is that it is defined somewhere on every neighborhood of (0,0). The reason for preferring this definition is that it let's us try to extend a function such as that given above to its full maximal domain of definition, in this case the whole plane. I.e. the given function has a well defined limit not only at (0,0) but also at every point of the line x= -y and all these points are removable discontinuities.

see e.g. page 36, chapter II.2 of Lang's Analysis I for this more precise definition of a limit of a function at a point which he requires there to be "adherent" to the domain, i.e. approachable as a limit of points of the domain. the same definition is given on page 126 of Advanced Calculus by Loomis and Sternberg.

An interesting question arises however in this context: suppose a function is undefined on an infinite subset of the closure of its domain, and suppose that the function does have a limit at each of those points. If we redefine the function to have as value the limit at each of those points, will the redefined function be continuous? Note that this is not entirely obvious since now we must check whether the new values of the newly defined function also converge to the other new values. i.e. each new value will of course be the limit of the values of the original function, but now we must also check that each new value is also a limit of those new values which are assigned to points near our point at which the function was originally undefined.

e.g. is it possible that a function could be undefined along the line x = -y, continuous elsewhere, and have a limit at each of the points of the line x = -y, but that the function defined on the whole plane by using those limiting values is not continuous? I.e. could it be that as we approach (0,0) along the line x = -y, that the limit values at those points do not approach the limit at (0,0?

I.e. is the existence of the limit of a continuous f, at every point of the closure of its domain, equivalent to the existence, as happens above, of a continuous extension to the whole closure of the domain? (It appears to me that it is but I did not write out the proof.)

 

[jambaugh]

[I deleted an earlier reply as I hadn't read your post (mathwonk) carefully enough.]

I think I see your point. Basically we can define limits on the boundary of the domain while restricting the limiting evaluations to the domain without any pathological weirdness. It's not terribly different from e.g. left and right limits in the 1 dim case, where we restrict the domain within the limit, or from limits at infinity.

 

[mathwonk]

that is an excellent example!

 

[Stephen Tashi]

If the two variable limit is defined as an iterated limit,
$$\lim_{(x,y)\to(0,0)} f(x,y) \equiv \lim_{x\to 0} \left[ \lim_{y\to 0} f(x,y) \right]$$

We should make it clear that the standard definition of a two variable limit is definitely not defined as an iterated limit. It would be better to say "If the two variable limit were defined as...", since this thought involves a fiction.

 

[Stephen Tashi]

the only restriction usually made on this definition is that the point (0,0) should be a limit point of the domain

The reason for preferring this definition is that it let's us try to extend a function such as that given above to its full maximal domain of definition, in this case the whole plane.

A different perspective is that such a definition keeps the topological approach to defining a limit from being ambiguous.

A topological attempt to define a limit would be to define the statement "$lim_{x \rightarrow p} f(x) = L$ " to mean:
For each open set $E$ containing $L$, there exists an open set $D$ containing $p$ such that $f(D - \{p\}) \subset E$.

Such a definition does not require that $p$ be in the domain of $f$. It also does not require that $f(D)$ be a non-empty set. So in the case where $p$ is not in the domain of $f$ and it is possible to find an open set $D$ containing $p$ such that $D$ is not in the domain of $f$ then $f(D - \{p\}) = \emptyset \subset E$ for any $E$ we choose. In this situation, we can pick $L$ arbitrarily and the above definition would say $lim_{x \rightarrow p} f(x) = L$.

For example, if the real valued function $f(x)$ is defined only on the domain $[0,1]$, we would have $lim_{x \rightarrow 5}f(x) = 13$, $lim_{x \rightarrow 5}f(x) = 14$, etc.

Since we want limits to be unique when they exist, the above attempt at a definition must be modified with conditions that prevent $f(D - \{p\})$ from being the empty set.

It's interesting that the unmodified definition given above translates to the epsilon-delta definition of limit typically presented in introductory calculus and the definition of a multi-variable limit often used in advanced calculus or even in some texts about metric spaces.

 

[mathwonk]

this is a good explanation for why it makes little sense to ask for the limit of a function at a point not in the closure of the domain, i.e. in that case the limit is never unique. there is nothing in it that applies only to the open set definition however. stephen's post makes clear that although one can give the definition of limit for points that are not in the closure of the domain, there is no good reason for doing so, as the limits are not unique and give no information at all about the function. the interesting question to me in this case is to provide a theorem corresponding to the one in the earlier case that allows a continuous extension of the function, i.e. supposing the function does extend continuously to the closure of it domain. then we know as stephen shows, that the function does have limits also at al points outside that closure, and in fact those limits can be arbitrarily chosen. can one then choose the limits at points outside that closure to give a continuous extension of the function to the whole plane? try the real line case first. (a hypothesis guaranteeing this is given in a famous lemma called maybe the tietze extension theorem.)

then take a special case; where the closure of the domain of the function is the closed unit disc in the plane, and assume the function is non negative, and bounded on that disc say by 1. then try to extend to a continuous function on the closed set inside some continuous curve C, so that the value on the boundary C of that set is constantly equal to zero. (picture the graph of your function on the unit circle as some closed curve in space, and try to deform that curve downward and outward to the plane z=0. e.g. try to project from a point on the z axis at height 2 say, to the plane z=0...does this generalize? ...)