Limit as (x,y) -> (0,0) ln(x^2+y^2)

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Homework Help Overview

The discussion revolves around evaluating the limit of the function ln(x^2 + y^2) as (x, y) approaches (0, 0). Participants explore different approaches to understand the behavior of the function near this point.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to prove the limit approaches -infinity by testing specific paths, such as setting x or y to zero. Others suggest using polar coordinates to analyze the limit more generally.

Discussion Status

Participants are actively exploring various methods to approach the limit, with some discussing the implications of using polar coordinates to simplify the problem. There is also a side discussion about a different limit involving three variables, where multiple approaches yield different results, prompting questions about the validity of those methods.

Contextual Notes

Some participants express uncertainty about the legality of setting variables equal to specific functions of another variable while evaluating limits. There is also mention of differing methods presented in textbooks, indicating a variety of approaches to limit evaluation in multivariable calculus.

theBEAST
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I know the answer is -infinity but I am not sure how to prove this. Would I just set x=0 and say that lim y-> ln(y^2) = -inf and do the same with y=0? But this would not prove it for all cases... Do I have to use the definition of the limit?
 
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The form (x^2+y^2) strongly suggests a parameterisation, yes?
 
Fightfish said:
The form (x^2+y^2) strongly suggests a parameterisation, yes?

What does that mean xD, I just started learning limits with two variables today and I found this question online and found it interesting. Something tells me my current knowledge is insufficient to solve this problem...
 
Ah, so you get to learn a new technique today:).

Let's use polar coordinates:
x = r cos \theta, y = r sin \theta
Then our limit becomes
\lim_{r \to 0} ln (r^2) = 2 \lim_{r \to 0} ln (r)​
You realize here that we have managed to remove the dependence on the direction of approach - ie our answer is independent of \theta.
 
Fightfish said:
Ah, so you get to learn a new technique today:).

Let's use polar coordinates:
x = r cos \theta, y = r sin \theta
Then our limit becomes
\lim_{r \to 0} ln (r^2) = 2 \lim_{r \to 0} ln (r)​
You realize here that we have managed to remove the dependence on the direction of approach - ie our answer is independent of \theta.

Wow this is a pretty cool way to solve limits, thanks!

I have one more question:
Does the limit as (x,y,z) -> (0,0,0) (xy+yz^2+xz^2)/(x^2+y^2+z^4) exist?

What I did was I set x,y = 0 and solved the limit as z -> 0 which gave me zero.

Next I set y=z^2 and x=z^2 and found the limit as z->0 which gave me 1. Therefore the limit does not exist.

Is this a correct method? I am not sure if it is legal to independently set y and z equal to z^2.


PS: the textbook did it differently they set y=x and z=0 and let the limit x -> 0 but I think there should be multiple ways to do this.
 
theBEAST said:
What I did was I set x,y = 0 and solved the limit as z -> 0 which gave me zero.
Next I set y=z^2 and x=z^2 and found the limit as z->0 which gave me 1. Therefore the limit does not exist.
Is this a correct method? I am not sure if it is legal to independently set y and z equal to z^2.
Yup. You can choose to approach the limit from any direction that you like. In your second case, you were approaching it from a parabolic path; that's fine.
theBEAST said:
PS: the textbook did it differently they set y=x and z=0 and let the limit x -> 0 but I think there should be multiple ways to do this.

Yes. A multivariable limit exists at a point only if the limit gives the same value no matter which path you take as you approach the point. To prove that a multivariable limit does not exist, it thus suffices to either show that 1) any two different paths give different limits or 2) there is a path for which the limit does not exist.

The tricky part in multivariable limits is to prove that a limit does exist, because you must show that the limit attains the same value for all possible paths of approach.
 

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