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Multivariable limits, how to show existence or non-existence

  1. Feb 21, 2012 #1
    1. The problem statement, all variables and given/known data
    lim(x,y)->(1,0) of ln((1+y^2)/(x^2+xy))


    2. Relevant equations


    3. The attempt at a solution

    Used two paths,
    x=1
    y=0
    both gave my lim=0
    so I tried x=rsin y=rcos, in attempt to use ε-δ to prove it.

    got to ln((1+r^2sin^2)/(r^2cos(cos+sin)))

    not sure where to go from here.
     
  2. jcsd
  3. Feb 21, 2012 #2
    Try going linear, with y = m(x - 1) instead.
     
  4. Feb 21, 2012 #3

    SammyS

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    I don't see that there's any difficulty as long as x→1 and y→0. ln(1/1) = 0
     
  5. Feb 21, 2012 #4
    My issue is in lecture my professor made it clear that finding any finite number of ways a function approached the same number did not prove that the lim was equal to that, in this case 0, because there are infinite number of ways (x,y) can approach the point. Does this make sense or did I misunderstand? Basically the only way he taught us to prove a lim existed was to use the ε-δ.
     
  6. Feb 21, 2012 #5

    SammyS

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    Yes, what your prof. said makes sense. I'm pretty sure that the functions that cause trouble are of indeterminate form, usually 0/0 . Very often the limit is being taken as (x,y)→(0,0) in which case using polar coordinates is often a big help.

    For the problem in this thread, you have neither 0/0, and (x,y)→(1,0) rather than (0,0).

    BTW: If the limit does not exist, then if you may be able to show that the limit is different along different paths.
     
  7. Feb 22, 2012 #6
    Well along the path x=y the limit blows up, 1/0, so then the limit does not exist?
     
  8. Feb 22, 2012 #7

    SammyS

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    The line y=x doesn't go through the point (1,0) .
     
  9. Feb 22, 2012 #8
    opps, ok. I think I solved it now.
    The limit exists and is equal to 0
    epsilon=r
    delta=r
    epsilon=delta.
     
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