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Limits of multiple variables along y=mx

  1. Oct 20, 2011 #1
    This is an assignment problem I have. Can't seem to figure it out. It has two parts, one to prove that the function approaches 0 where y=mx and and one where the function approaches 0 from y=kx2...

    f(x,y) = (x3y)/(2x6+y2)

    I've attempted both parts and get stuck on what seems like the same issue. Basically I've substituted y=mx for y. Giving:

    Please assume that lim (x,mx)→(0,0) is at the beginning of these workings.
    lim f(x,mx) = x3(mx)/(2x6+(mx)2)
    lim f(x,mx) = mx4/[(x2(2x4+m2)]
    lim f(x,mx) = mx2/(2x4+m2)

    I suppose I could try L'Hopital at this point, but it doesn't seem to help I think the 3rd derivative leaves you with 0/0.

    Could I get some direction, please?

    Thanks prior!
     
    Last edited: Oct 20, 2011
  2. jcsd
  3. Oct 20, 2011 #2

    SammyS

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    You have y = mx, so that if x→0, then mx→0 . This changes taking the limit as (x,y)→(0,0) to taking the limit as x→0.

    For your problem:
    If y = mx, then lim (x,y)→(0,0) f(x,y)
    becomes, lim x→0 f(x,mx), where [itex]\displaystyle f(x,\,mx)=\frac{x^3(mx)}{2x^6+(mx)^2}=\frac{mx^4}{x^2(2x^4+m^2)}[/itex]
     
  4. Oct 20, 2011 #3
    Thanks for your quick reply. That definitely helped me understand. I was able to answer both parts of the question with your help. The third part says to show that lim along y=x3 does not exist.

    Now, I figured it to be 1/3, but I think the book is implying that the limit does not exist because the first two paths worked, but the third was different, therefore the limit does not exist. Would you agree?

    Thanks again for your help! I spent way too much time dinkin' around with that problem...
     
  5. Oct 20, 2011 #4

    SammyS

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    That's the idea.
     
  6. Oct 21, 2011 #5

    lanedance

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    why use L'Hopital? the limit you show is not of type 0/0 as x->0
     
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