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Homework Help: As (a,b)->(0,0) limit (a,b)/(a^2+b^2) exists

  1. Sep 21, 2006 #1
    As (a,b)-->(0,0) limit (a,b)/(a^2+b^2) exists

    I'm stuck on epsilon delta argument for (a,b)-->(0,0) limit (ab)/(a^2+b^2) exists. HELP
     
    Last edited: Sep 21, 2006
  2. jcsd
  3. Sep 21, 2006 #2

    HallsofIvy

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    For problems like this, I recommend converting to polar coordinates:

    [itex]a= r cos(\theta) b= r sin(\theta)[/itex] so
    [tex]\frac{ab}{a^2+ b^2}= \frac{r^2 sin(\theta)cos(\theta)}{r^2}[/tex]

    [tex]= cos(\theta)sin(\theta)[/tex]
    The distance from (0,0) to (a,b) is just r.
    ( Hmm, I forsee a serious problem in proving that limit exists!)

    In the title you said (a,b)/(a^2+ b^2) which I would interpret as the vetor
    [tex](\frac{a}{a^2+ b^2},\frac{b}{a^2+b^2})[/tex]
    In polar coordinates that is
    [tex](\frac{cos(\theta)}{r}, \frac{sin(\theta)}{r})[/tex]
    Now it is easy to see that each component is less than 1/r and, again, r measures the distance from (0,0) to the point (a,b).
     
  4. Sep 21, 2006 #3
    umm... again sorry but
    function is (ab)/(a^2+b^2) comma shouldn't be there...
    sorry
     
  5. Sep 22, 2006 #4
    Yes, you have seen it. actually the limit doesn`t exist,because the parameter [tex]\theta[/tex] in your equation isn`t a constant.
     
  6. Sep 22, 2006 #5

    HallsofIvy

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    Well, in that case, there is no "epsilon-delta" argument that the limit exists for a very good reason. The limit does not exist.
     
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