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- Thread starter theBEAST
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The form (x^2+y^2) strongly suggests a parameterisation, yes?

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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...

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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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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

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Yup. You can choose to approach the limit fromWhat 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.

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

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