Best way of evaluating limits of multi variable functions

[VietDao29]

'A point z is a limit point of a set S (real or complex) if any neighbourhood of z contains a point of S, other than z itself.'
Hence only one point (meaning one path into the origin) is enough to prove that a limit exists.
So with your example, even if the function is undefined going along the x or y axis, it does not mean that another path cannot approach the origin legitamatly. Therefore there might be a point around the neighbourhood of the origin that is an element of the function \lim_{(x, \ y) \rightarrow (0, \ 0)} |xy| \ln|xy|

Arghhh,... uhmmm, I think you should read my post carefully.
I take that example from this thread. Let's see if you can get it. The limit does not exist, not 0.
This limit:
\lim_{(x, \ y) \rightarrow (0, \ 0)} (|x| + |y|) \ln(|x| + |y|) = 0
Can you see the difference?

One suspician I have with using polar coords is that there might exist some complicated function which have a limit but can only be known by a very complicated and weird path into the origin. The polar coords is not able to pick it up and nor are the other tools like the sandwich theorem be useful.

As I told you before, you should be flexible. Some function like sine, and cosine, logarithm, blah, blah, blah... then you should use something else, polar coord won't help these cases, I think.
Now look at the thread I've shown you to see if you can get it. :)

 

[pivoxa15]

Uhmm, this is not quite correct... :)

3. There is no neighbourhood of (0, 0) that the function is defined for every x, and y in that neighbourhood.
------------------
The limit:
\lim_{(x, \ y) \rightarrow (0, \ 0)} |xy| \ln|xy|
does not exist. Since the function is not defined along the x, or y-axis (number 3).
The limit still does not exist even if it does evaluate to 0 as you approach it along the x = ky path, or y = kx (k is not 0), or even in a parabolic path like y = x² (Or whatever path you like )...
However, the limit \lim_{(x, \ y) \rightarrow (0, \ 0)} (|x| + |y|) \ln (|x| + |y|) = 0 (explained in the previous post). Notice that the function is defined for every pair of (x, y), but (0, 0). Right?

How sure are you have criteria 3? Where did you get the information from?

Why do you need a whole defined (by the function) neibourhood around the origin in order for the limit to exist? Why not just allow at least one point on any neighbourhood around the origin to exist?

 

[VietDao29]

How sure are you have criteria 3? Where did you get the information from?

Why do you need a whole defined (by the function) neibourhood around the origin in order for the limit to exist? Why not just allow at least one point on any neighbourhood around the origin to exist?

By the definition of limit of 2 variable function, we haveL
\lim_{\substack{x \rightarrow p \\ y \rightarrow q}} f(x, \ y) = L
if and only if:
For every ε > 0 there exists a δ > 0 such that for all real numbers x, y with 0 < ||(x,y)-(p,q)|| < δ, we have |f(x,y)-L| < ε
Where ||(x,y)-(p,q)|| represents the Euclidean distance.
Now, let's look at the bolded part:
0 < ||(x,y)-(p,q)|| < δ, that means that there must exist a circular region with radius δ (excluding the center (p, q)), such that for every point (x, y) in that region, we have: |f(x,y)-L| < ε.
So if there happens to be some point (x, y) in that circular region that the function f(x, y) is not defined, then is |f(x,y)-L| defined?
No, right? So can we say that: for every number x, y in that region, the following statement is true |f(x,y)-L| < ε? Again, it's no, right?
So again, if there exists no circular region arround (p, q) (excluding the point (p, q) itself), that for every pair of number (x, y) the function f(x, y) is defined, then the limit does not exist (as it violates the definition).
Now can you get it? :)

 

[pivoxa15]

I get it for this example. Can you prove why polar coords work and why sometimes it does not work?

 

[VietDao29]

I get it for this example. Can you prove why polar coords work and why sometimes it does not work?

It will always work. But as I said earlier, sometimes it will take more time to solve a problem in polar coord, than in some way else. Consider this example:
\lim_{(x , \ y) \rightarrow (0, \ 0)} \frac{1 + x ^ 2 + y ^ 2}{y ^ 2} (1 - \cos y)
Now since we have:
\lim_{y \rightarrow 0} \frac{1 - \cos y}{y ^ 2} = \frac{1}{2}. The whole expression will tend to:
\lim_{(x , \ y) \rightarrow (0, \ 0)} \frac{1 + x ^ 2 + y ^ 2}{y ^ 2} (1 - \cos y) = \frac{1}{2} (1 + 0 ^ 2 + 0 ^ 2) = \frac{1}{2}.
But if you change to polar coordinate:
\lim_{(x , \ y) \rightarrow (0, \ 0)} \frac{1 + x ^ 2 + y ^ 2}{y ^ 2} (1 - \cos y) = \lim_{r \rightarrow 0} \frac{1 + r ^ 2}{r ^ 2 \sin ^ 2 \theta} (1 - \cos (r \sin \theta)). This will take more time to solve, and look much more complicated than the previous way I've shown you right?
So overall, if there are some functions like sine, cosine, tangent, cotangent, exponential, blah blah blah, don't use polar-coord.
Only use polar coordinate when you encounter some fraction, of which both numerator, and denominator are polynomials.
Can you get this? :)

 

[pivoxa15]

It will always work. But as I said earlier, sometimes it will take more time to solve a problem in polar coord, than in some way else. Consider this example:
\lim_{(x , \ y) \rightarrow (0, \ 0)} \frac{1 + x ^ 2 + y ^ 2}{y ^ 2} (1 - \cos y)
Now since we have:
\lim_{y \rightarrow 0} \frac{1 - \cos y}{y ^ 2} = \frac{1}{2}. The whole expression will tend to:
\lim_{(x , \ y) \rightarrow (0, \ 0)} \frac{1 + x ^ 2 + y ^ 2}{y ^ 2} (1 - \cos y) = \frac{1}{2} (1 + 0 ^ 2 + 0 ^ 2) = \frac{1}{2}.
But if you change to polar coordinate:
\lim_{(x , \ y) \rightarrow (0, \ 0)} \frac{1 + x ^ 2 + y ^ 2}{y ^ 2} (1 - \cos y) = \lim_{r \rightarrow 0} \frac{1 + r ^ 2}{r ^ 2 \sin ^ 2 \theta} (1 - \cos (r \sin \theta)). This will take more time to solve, and look much more complicated than the previous way I've shown you right?
So overall, if there are some functions like sine, cosine, tangent, cotangent, exponential, blah blah blah, don't use polar-coord.
Only use polar coordinate when you encounter some fraction, of which both numerator, and denominator are polynomials.
Can you get this? :)

That is a nice example. But efficiencies aside, can you prove that if a limit in polar coords is found than it will be exactly the same as the limit when not using polar coords and using another method?

 

[VietDao29]

That is a nice example. But efficiencies aside, can you prove that if a limit in polar coords is found than it will be exactly the same as the limit when not using polar coords and using another method?

Uhmm, have you tried to do it yourself?
As r tends to 0, r \sin \theta \rightarrow 0
So:
\lim_{r \rightarrow 0} \frac{1 - \cos (r \sin \theta)}{(r \sin \theta) ^ 2} = \frac{1}{2}. And the whole limit is:
\lim_{r \rightarrow 0} \frac{(1 + r ^ 2) (1 - \cos (r \sin \theta))}{(r \sin \theta) ^ 2} = \frac{1}{2}.
Can you get it? :)

 

[pivoxa15]

Uhmm, have you tried to do it yourself?
As r tends to 0, r \sin \theta \rightarrow 0
So:
\lim_{r \rightarrow 0} \frac{1 - \cos (r \sin \theta)}{(r \sin \theta) ^ 2} = \frac{1}{2}. And the whole limit is:
\lim_{r \rightarrow 0} \frac{(1 + r ^ 2) (1 - \cos (r \sin \theta))}{(r \sin \theta) ^ 2} = \frac{1}{2}.
Can you get it? :)

Yes I have tried doing it and it does come to 1/2 by using L'ptol's rule which involved many long product differentiations.

What I tried to say in the previous post was not directly focused on your specific example but situations in general. Is it enough to show that the evaluated limit found by using polar coords as r->0 will everytime turn out to be the limit of the f(x,y) function when x,y->0?

Does it bother you that when using polar coords we are really fixing a point through theta and let the radius of the straight line go to 0? It means that we are really going into the origin via a straight line. There might exist a really complicated function which has a limit A but its polar form gives a limit B or is undefined. Hence by showing that a function in polar form possesses a limit in the origin does not have to imply that the function in cartesian coords has the same limit in the origin.

 

[VietDao29]

Does it bother you that when using polar coords we are really fixing a point through theta and let the radius of the straight line go to 0? It means that we are really going into the origin via a straight line. There might exist a really complicated function which has a limit A but its polar form gives a limit B or is undefined. Hence by showing that a function in polar form possesses a limit in the origin does not have to imply that the function in cartesian coords has the same limit in the origin.

Nah, you've misinterpreted my points, \theta needs not to be a constant, it can be a function of r, or whatever. That's why the limit in my example 2 does not exist. You can skim through the 3 examples again to get what I am saying.
If the limit as r tends to 0 does not depend on \theta, then the limit exists, otherwise, the limit does not exist.
Is there anything else unclear? :)

 

[pivoxa15]

So you are suggesting that it does not matter which coordinate system you use, as long as it can evaluate an unique finite number than the limit for this particular function exists and it is that one. For example you can't get a different limit value by switching to another coordinate system.

Could you prove it?

 

[VietDao29]

So you are suggesting that it does not matter which coordinate system you use, as long as it can evaluate an unique finite number than the limit for this particular function exists and it is that one. For example you can't get a different limit value by switching to another coordinate system.

Could you prove it?

You can think about it this way.
Say you want to evaluate the limit of a 2 variable function, as (x, y) -> (0, 0)
For every point (x, y) in Cartesean coordinate, it can be written in polar coordinate as (r, \ \theta).
Now as (x, y) -> (0, 0), it means that r -> 0, right (Since r = \sqrt{x ^ 2 + y ^ 2})? And \theta can take whatever value, right?
Can you get this?
And if the limit for the new expression as r -> 0, just depends on r, and independent of \theta. That means as (x, y) -> (0, 0), no matter which path we go along, we will obtain only one limit, right? (Since \theta can be chosen arbitrarily, and that \theta does not need to be a constant, it can also be a function of r, and \theta does not affect the outcome of the result).
Can you get this? :)

 

[Pradeep.pala]

help me with this

Can anyone help me with this

let f(x,y)= ((x^2) * (y^2) ) / (x^2 + y^2) if ε>0 find d(delta) so that
0< \sqrt{x ^ 2 + y ^ 2} < d implies | f(x,y) | < ε