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

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  • #1
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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?
 

Answers and Replies

  • #2
954
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The form (x^2+y^2) strongly suggests a parameterisation, yes?
 
  • #3
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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...
 
  • #4
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Ah, so you get to learn a new technique today:).

Let's use polar coordinates:
[tex]x = r cos \theta, y = r sin \theta[/tex]
Then our limit becomes
[tex]\lim_{r \to 0} ln (r^2) = 2 \lim_{r \to 0} ln (r)[/tex]​
You realise here that we have managed to remove the dependence on the direction of approach - ie our answer is independent of [itex]\theta[/itex].
 
  • #5
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Ah, so you get to learn a new technique today:).

Let's use polar coordinates:
[tex]x = r cos \theta, y = r sin \theta[/tex]
Then our limit becomes
[tex]\lim_{r \to 0} ln (r^2) = 2 \lim_{r \to 0} ln (r)[/tex]​
You realise here that we have managed to remove the dependence on the direction of approach - ie our answer is independent of [itex]\theta[/itex].
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.
 
  • #6
954
117
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.
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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