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Limits with three variables

  1. Oct 13, 2008 #1
    1. The problem statement, all variables and given/known data
    lim as (x,y,z) ->0,0,0 of xyz/(x^2+y^2+z^2) it says (HINT: use spherical coordinates)


    2. Relevant equations



    3. The attempt at a solution

    I tried putting it in spherical coordinates but it didn't seem to help. I ended up getting pcos(theta)sin(psi)p(sin(theta)sin(psi))pcos(theta)/(p^2cos^2(theta)sin^2(psi)+p^2sin^2(psi)sin^2(theta)+p^2cos^2(psi))

    I don't know what I can do with this messy thing or even why theywould tell me to use it just looks like it made it worse. By the way my teacher said we must use spherical coordinates or we will receive no credit.
     
  2. jcsd
  3. Oct 13, 2008 #2
    Are you sure you don't have a better expression for your denominator using the formulas relating spherical and cartesian coordinates?

    Then, if (x,y,z)->0, what is/are going to what limit in spherical?
     
  4. Oct 13, 2008 #3
    oh i see you could put the denominator as p^2 correct? and then the top would go to 0 as (x,y,z) go to zero?
     
  5. Oct 13, 2008 #4

    Dick

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    The 'top' going to zero is not enough. The top has to go to zero faster than p^2. Does it?
     
  6. Oct 13, 2008 #5
    actually wouldn't the top have a p^3 term so it would go to 0 faster then the bottom?
     
  7. Oct 13, 2008 #6

    Dick

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    What's the power of p on the top?
     
  8. Oct 14, 2008 #7
    it's to the third power right so you end up with a 1 in the denominator and a p in the numerator?
     
  9. Oct 14, 2008 #8

    Dick

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    Yes, and p goes to zero, right? What about the trig functions? They are bounded, right?
     
  10. Oct 14, 2008 #9
    yea they are bounded between -1 and 1 and p goes to 0 so the top becomes 0 and the limit is 0
     
  11. Oct 14, 2008 #10

    Dick

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    Right. The quotient bounded between -p and p and p goes to zero. Well done.
     
  12. Oct 14, 2008 #11
    thanks for the help
     
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