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Multivariable Calculus: Limits

  1. Sep 16, 2007 #1
    1) lim [x(y^2)] / (x^2 + y^2)
    (x,y)->(0,0)
    Find the value of the given limit, if it exists.

    Using polar coordinates, set x = r cos(theta), y = r sin(theta)
    Then, the given limit = lim [r cos(theta) r^2 sin^2(theta)] / r^2
    r->0
    = lim r [cos(theta) sin^(theta)]
    r->0
    = 0 since cos(theta) sin^(theta)<=1, i.e. bounded

    If I found that the limit is equal to 0 this way, can I conclude immediately that the original limit is 0 too?

    What I believe is that for the part using polar coordinates r->0, it seems that it's appraoching the origin through straight lines paths ONLY, however my textbook says that "for the limit to exist, we must get the same result no matter which of the infinite number of paths is chosen"


    Thanks for answering!
     
  2. jcsd
  3. Sep 16, 2007 #2
    If I found that the limit is equal to 0 this way, can I conclude immediately that the original limit is 0 too?

    yes you can


    i believe when you switch to polar


    and after simplification

    plug in your r

    if your equation is not a function of theta, then that is the limit


    if your equation is still a function of theta, then the limit DNE



    and you can also use lhopitals rule once you switch to polar
    if the conditions are met
     
    Last edited: Sep 16, 2007
  4. Sep 16, 2007 #3

    HallsofIvy

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    No, you are not approaching "through straight lines only" because you are not assuming that [itex]\theta[/itex] is a constant. You need to show that the limit goes to 0 as r goes to 0, no matter what [itex]\theta[/itex] is. The real point is that, in polar coordinates, r measures exactly the distance to the origin, irrespective of [itex]\theta[/itex]- and that is what must be made "less than [itex]\delta[/itex]".
     
  5. Sep 16, 2007 #4
    "You need to show that the limit goes to 0 as r goes to 0, no matter what theta is" <---but this covers only the paths of all the straight lines through the origin, how about in the paths of parabolas and cube root function through the origin, etc?
     
  6. Sep 16, 2007 #5
    Because theta can be a function of r...
     
  7. Sep 16, 2007 #6
    But how does this help? Sorry, I am not getting it...
     
  8. Sep 16, 2007 #7
    It's a straight line iff theta is a constant. But the proof allows theta to vary arbitrarily, therefore, you're allowed curves.
     
  9. Sep 16, 2007 #8
    How can it be a curve if theta can vary?
     
  10. Sep 16, 2007 #9
    Consider the equation [tex]\theta = r[/tex]. Plot it. Is it not a curve?
     
  11. Sep 16, 2007 #10

    arildno

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    It seems you are unsure of what theta really is.

    So I have to ask:

    What, precisely, does the variable theta measure?
     
  12. Sep 16, 2007 #11
    theta is a counterclockwise measure of "angle" from the positive x-axis, this is what makes me think of r->0 as (x,y) approaching (0,0) through straight lines only
     
  13. Sep 16, 2007 #12
    Did you try and plot the function \theta = r as I suggested? It is a straight line?
     
  14. Sep 16, 2007 #13
    Do you mean [r cos(r) r^2 sin^2(r)] / r^2 ?

    By the way, we are taking the limit r->0, shouldn't theta be fixed?
     
  15. Sep 17, 2007 #14

    arildno

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    Why?
    Any particular POINT can be uniquely specified by its distance from the origin, and the angle the line segment between the origin and the point makes with the positive x-axis.

    By no means does this entail that all the points a particular curve consists of make the same angle to the positive x-axis with their respective line segments.
     
  16. Sep 17, 2007 #15
    No! That's the point! We DON'T have to fix theta!
     
  17. Sep 17, 2007 #16

    arildno

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    You can approach the origin by way of a straight line, but also by spiralling yourself inwards. And in many other ways as well.
    Only if you approach the originalong a straight line will theta be constant.
     
  18. Sep 18, 2007 #17
    If theta is allowed to vary, will r->0 using polar coordinates cover ALL paths for which (x,y)->(0,0) ?
     
  19. Sep 18, 2007 #18
  20. Sep 18, 2007 #19

    HallsofIvy

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    "Paths", in polar coordinates, depend only on r and [itex]\theta[/itex]. If r ranges from, say, 1 to 0, while [itex]\theta[/itex] is allowed to have any value, then, yes, it covers all paths. More to the point is that in polar coordinates, r alone measures the distance from the origin which is what you want: ||(x,y)- (0,0)||= r< [itex]\delta[/itex].
    As long as the limit, as r goes to 0, is a number, that is, independent of [itex]\theta[/itex] that number will be the limit.
     
    Last edited: Sep 20, 2007
  21. Sep 18, 2007 #20
    I think part of the confusion might be this: a path near the origin, if it goes through it, can be well approximated by a straight line. Unfortunately, whilst that's intuitively true, analysis can always come back to bite us:

    [tex]\theta = \pi sin(1/r)[/tex]
     
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