How to tell if a multivariable limit really exists?

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in multivariable functions you can say that a limit exist proving it by definition. You only use different paths to see if a limit doesn't exist (case that you get different values), or to find a candidate to be the limit and use it to proving that by definition.

You don't have a way to see exactly which paths you have to use. You can use the lines x=0, y=mx and then all depends on your creativity.

Sometimes can be useful to plot some level sets of the function, and analize if exist a trivial path which aproximates the point in a special way that can give a diferent value of the limit.

I hope it had been helpful.

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NeptaliD
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NeptaliD

Updated on September 03, 2020

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  • NeptaliD
    NeptaliD about 3 years

    Hi i'm having this doubt about this multivariable limit

    $$ \lim\limits_{(x,y)\to(0,0)} \frac{2x^2y}{x^4+y^2} $$

    If I use these two paths like let $x = y^3$ and $y=x$ I get that the limit is going to be 0, however if I add this other path where $x=\sqrt y$ the limit is going to be 1, and that tells me the limit doesn't exist, but what if I just used the first ones? while doing the proof by definition shouldn't this be refuted? cause I don't know if i'm applying this the wrong way and ending up with a valid limit when I shouldn't, how do you guys know exactly which paths you can use to say that the limit doesn't exists? is there like a rule or something similar? Thanks in advance!

  • NeptaliD
    NeptaliD about 7 years
    Exactly, in this case i used the first two paths as i earlier mentioned I got 0, then by definition that's the value i used then I got to the point where the absolute value of the the function is < than Epsilon and I thought I proved it, but then I noticed afterwards because a friend told me if you used that path where $x = \sqrt y$ the limit doesn't exists cause is different as the other two paths I already used, but then again the main question is does the definition tells me if it really exists? even wolfram alpha says this limit is 0 however in leithold's 7th edition says it doesn't exists
  • DN_Euler
    DN_Euler about 7 years
    Yes, limit by definition always work. Maybe you do something wrong when you try to proof it? I try to solve it and using the path x=sqrt(y) the limit goes to 1, but using the path y=mx goes to 0. So, the limit doesn't exist. Don't trust always on Wolfram Alpha.
  • NeptaliD
    NeptaliD about 7 years
    Alright thanks so much yeah i'm gonna try and see what's the failure with the definition I tried
  • DN_Euler
    DN_Euler about 7 years
    Excellent, good luck!
  • NeptaliD
    NeptaliD about 7 years
    I get that, however how do you prove that the limit doesn't exist by definition? i don't see what i did wrong while proving it was 0, and it is a wrong answer
  • Alex Ortiz
    Alex Ortiz about 7 years
    @NeptaliD, I have updated my answer to address that question.