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I Multivariable limits

  1. Jul 29, 2016 #1
    I have found that multivariable limits are harder to find and/or prove that something exists.
    Do you have any recommendations, given questions like "find(if exists) the limit....".

    For example, I have no idea how to even start thinking about the following limit(if it exists or not, and if it does, what is the value):
    [tex] \lim_{\textbf{x}\rightarrow 0} \frac{e^{|\textbf{x}|^2}-1}{|\textbf{x}|^2 +x_1^2x_2+x_2^2x_3} [/tex]

    where [tex] \textbf{x}=(x_1,x_2,x_3)[/tex]

    I would like to get any help for dealing with limits such as the aforementioned and others.

    Thank you.
     
    Last edited: Jul 29, 2016
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  3. Jul 29, 2016 #2

    micromass

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    First things I would do:
    1) Plot the function with a graph tool. This way you might see if the limit exists or not.
    2) If the limit exists, then if ##\mathbf{x}_n\rightarrow \mathbf{x}##, then ##f(\mathbf{x}_n)\rightarrow \text{limit}##. So choose some simple sequences and see if they all yield the same limiting value.

    Also, I'm not sure what ##\mathbf{x}\rightarrow \infty## means, can you define it?
     
  4. Jul 29, 2016 #3

    BvU

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    Hi,

    not a very good example: the ##e^{\vec x^2}## trumps everything. No limit.

    In general:
    focus on the heavies: the ##-1## in the numerator can be ignored.
    if sensible, divide numerator and denominator by the heaviest term and see what's left over.
    If what's left tends to 0/0 see which goes to 0 the fastest:
    use taylor series
    use l'Hopital's rule
    google 'limit theorems' to find more !
     
  5. Jul 29, 2016 #4

    mfb

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    It could still go to infinity, which is not a limit in the real numbers but still some sort of limit. It does not, but that needs further analysis here.
     
  6. Jul 29, 2016 #5

    micromass

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    Am I the only one who doesn't know what ##\mathbb{x}\rightarrow \infty## means? How are you guys interpreting this?
     
  7. Jul 29, 2016 #6
    Oops, it should go to 0. Sorry.
    What would you do if you get this question on an exam or in a situation when you can't use a graphical tool? how would you proceed?
    Probably the reason why the example isn't good is because ##\mathbf{x}\rightarrow \infty## should be ##\mathbf{x}\rightarrow 0##.
    How do you use l'Hopital's rule in multivariable calculus?

    Thank you for your answers.
     
  8. Jul 29, 2016 #7

    mfb

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    If ##\lim_{a \to \infty} \sup_{|x|=a} f(x) = \lim_{a \to \infty} \inf_{|x|=a} f(x) ## (assuming those things are well-defined), it makes sense to call this value "limit" I think.

    For x -> 0 I would multiply numerator and denominator with the same thing to make the denominator nicely convergent (with a finite limit), which allows to simplify the expression.
     
  9. Jul 29, 2016 #8

    BvU

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    Back to the example that was slightly modified to become more interesting. I would interpret ##\vec x\downarrow \vec 0## as: going to ##(0,0,0)## from any possible direction. (Must admit I never had any of these vector limits on an exam -- at least I don't remember :smile:).

    Taylor series is good for the numerator: ##\exp (x_1^2 +x_2^2+x_3^2) - 1 = (x_1^2 +x_2^2+x_3^2) + \mathcal O\left ( (x_1^2 +x_2^2+x_3^2)^2 \; \right ) ##.
    Divide numerator and denominator by ## (x_1^2 +x_2^2+x_3^2) ## and get (1 + small stuff) / (1 + small stuff) ##\Rightarrow ## limit exists and is 1.
     
  10. Jul 29, 2016 #9

    micromass

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    It makes sense, and this is the likely interpretation if you get something like this. But it is not the only possible interpretation. With ##\mathbf{x}\rightarrow 0## it is very unambigious. With "##\mathbf{x}\rightarrow \infty##, I think it should be defined first. But anyway, the problem has an error so it's not important.
     
    Last edited by a moderator: Jul 29, 2016
  11. Jul 29, 2016 #10
    Thank you. This is really becomes easy with Taylor series. I think I should use it more often in these situations.

    I will collect your advice and try to do some more exercises. Still, if you have more suggestions, I would be grateful to hear them.

    Thank you all.
     
  12. Jul 29, 2016 #11

    mfb

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    "From any possible direction" works only if you include curved "directions". There are functions where all straight lines lead to 1-dimensional functions that converge nicely, but the actual two-dimensional function does not converge. One example:
    $$f(x) = \frac{xy}{x^2+y}$$

    As bonus, it makes the three-dimensional limit one-dimensional, as we have to care about |x| only.
     
  13. Jul 29, 2016 #12

    fresh_42

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    I didn't know either. At latest when it comes to differentials such a sloppy notation will heavily strike back.
     
  14. Jul 29, 2016 #13
    I take this opportunity to make a basic question.
    there is some method to determine if the limit exists?, and not try all possible ways to reach the indeterminate point.
    example:
    upload_2016-7-29_23-28-52.png
    the limit is 0 but the method is try the path of x=0, y=0, y=mx , y=mx^2+nx.
    there any way to know if the limit exists or not exist that does not involve check all the paths like a computer ?
     
  15. Jul 30, 2016 #14
    My first thought when I saw this limit was moving to polar coordinates, because it has terms that look like ##x^2+y^2##. So I just added in the denominator ##+4x^2-4x^2(=0)## to get something with the ##5y^2##. Then I arrived to ##r\cdot ##(some finite number) which goes to 0 as ##r## goes to 0.
     
  16. Jul 30, 2016 #15

    mfb

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    There is an infinite set of paths, you cannot check them all.

    Going to radial coordinates is often a good approach.
     
  17. Jul 30, 2016 #16
    radical coordinates is good, ok.
    thanks =D
     
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