Best way of evaluating limits of multi variable functions

[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 independant 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