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CalcIII limit problem

  1. Sep 22, 2009 #1
    hi.
    Spse you want to find limf(x,y) as (x,y)->0.
    You can use polar coordinates and get
    limf(rcost,rsint) as r->0. And these limits are same.
    Now, i initially thought that
    1)if limf(rcost,rsint)is independent of t,then limit
    exists and is equal tolimf(x,y) as (x,y)->0.
    2)If it depends on t, then limit DNE.
    But, i did this limits using these methods and got different answers:
    limit(xy^{4}/x^{2}+y^{8}) as (x,y)->0.
    So if you try x=y^{4} path, you get 0.5
    y=0 path, you get 0. So DNE
    if you convert into polar, you get zero.

    Where is the mistake??
    Thanks in advance
     
  2. jcsd
  3. Sep 23, 2009 #2
    I cannot tell you whether your limit exists or not, but I can tell you why you get different answers. In polar coordinates, when you take the limit as r -> 0 and t is fixed, you are "only" checking the limit along straight lines. It is equivalent to checking x = m y only.

    I don't know whether or not that's sufficient for the limit to exist. But when you graph your function in close neighborhoods of (0,0), the limit seems to be (0,0) indeed.
     
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