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Two path test

  1. Sep 29, 2017 #1

    yecko

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    1. The problem statement, all variables and given/known data
    Wx0tkw8.jpg
    https://i.imgur.com/Wx0tkw8.jpg
    (example 3b)
    2. Relevant equations
    limit of multivariable equation
    Two path test

    3. The attempt at a solution
    My instructor told us that we should only use two paths test only if we predict there isn't any limit... but how can we 'sense' there aren't limit?? (take e.g. 3b...)
    moreover, where are the two paths come from? (I don't feel like y=0 and y=-x^2 have any connection to the required function...)

    any help would be appreciated~~~
    thank you very much!!!
     
  2. jcsd
  3. Sep 29, 2017 #2

    FactChecker

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    In 3b, the x and y have different powers inside the sin(). Everything else is the same for them. So you can suspect that the limits for the path y≡0, x→0 and the path x≡0, y→0, may be different.

    Picking the path where y = -x2 is clever. It makes the numerator always 0 on that path, so that limit on that path is 0.
     
  4. Sep 29, 2017 #3

    yecko

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    What does it imply for everything the same??
    (indeed y is to power 1 in numerator and denominator y is to power 2...)
    as they are both tends to 0, why are they may be different??

    Thank you!
     
  5. Sep 29, 2017 #4

    FactChecker

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    If there is one significant difference and everything else is identical, you can more easily determine what the effect of that one difference will be.
    That is not true. Even though both numerator and denominator go to 0, they go to zero at different rates. The limits will depend on the rates. I don't know what techniques you have to analyse that. If you know the "small angle approximation", you can use that here.
     
  6. Sep 29, 2017 #5

    yecko

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    oh! I get it! use l'hospitals rule, right?
    I am still doubting where y=-x^2 comes from... or it is just any random functions do?

    thanks
     
  7. Sep 29, 2017 #6

    FactChecker

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    Yes. That will work. Or you can use the small angle approximation sin(θ) ≈ θ if θ ≈ 0.
    It makes x2 + y ≡ 0 on that path. So the numerator is 0 on that path.
     
  8. Oct 2, 2017 #7

    scottdave

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    One thing to note. Sine function is odd, while a parabola is even. But sine of x^2 is going to be even, since its argument is always positive. You could use that knowledge to help predict that it may not work from 2 different paths. Set y = 0 then consider x -> 0. Then do the same with x=0 and y->0. How do they differ?
     
  9. Oct 3, 2017 #8

    Ray Vickson

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    For question (b), change to polar coordinates ##x = r \cos \theta## and ##y = r \sin \theta##.
     
  10. Oct 6, 2017 #9

    haruspex

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    My guess is that your instructor means this:
    To prove it does not converge to the same value on all paths, you should try to find two paths with different limits; to prove it does converge on all paths you need to use a more comprehensive approach - no matter how many specific paths you test you will still not have proved it.
    At first, you may not know which is true, so be prepared to alternate. If the limit is the same on a few interesting looking paths then try to prove the general result; when that fails, try to understand why it eludes you, as that can give the clue as to what pathological path to try.
     
  11. Oct 6, 2017 #10

    FactChecker

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    To be suspicious of there not being a limit, consider any point where the denominator becomes 0. If the numerator and/or denominator approaches 0 at different rates depending on the path (proportional to different powers of the variables), then you should look hard at that.
     
  12. Oct 6, 2017 #11

    Ray Vickson

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    The suggestion in #8 will show immediately what is happening.
     
  13. Oct 6, 2017 #12

    yecko

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    Thank you very much for all your replies!
     
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