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A difficult limited

  1. Apr 24, 2008 #1
    When we think a value an end in the compound functions :
    We remember that we in front of number from not a limited from the paths
     

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  2. jcsd
  3. Apr 24, 2008 #2

    Dick

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    We remember that we in back of path from not a limited from the numbers. Is that wrong? I really can't understand you. Sorry.
     
  4. Apr 25, 2008 #3
    He is not at the end a presence
    Why ؟
    The previous way the others is suitable ? why ?...
    Look please:
     

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  5. Apr 25, 2008 #4

    Dick

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    I will. When it's approved. I think that gives me time to get a good night's sleep.
     
  6. Apr 25, 2008 #5

    Dick

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    Now I don't understand what your picture is trying to tell me either. The proof of the limit in your first post is just fine.
     
  7. Apr 26, 2008 #6

    I did not understand your intention ?
    I do not have a confirmation of the response
    But look at the next example:
    I apologize
    Second : on the axis z = y^2
     

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    Last edited: Apr 26, 2008
  8. Apr 26, 2008 #7

    HallsofIvy

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    You have now posted 3 different pictures that appear to involve three different problems. As Dick told you before, the first is correct. This third one is also correct- except for spelling: since you get two different limits by approaching (0,0) along two different paths, the limit does not exist.

    I'm not sure what the second picture was supposed to represent. But one thing you should be careful about: The line y= 2x, for example, in R2, is not in the complex plane.
     
  9. Apr 26, 2008 #8

    My answer of the first question :
    is correct or a mistake?


    remember:
    x=(z+z')/2
    y=(z-z')/2i
    z'=z par
     

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  10. Apr 26, 2008 #9

    Dick

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    lim(z->0) is not the same as lim(x->0) or lim(y->0). It's lim(x->0 AND y->0).
     
  11. Apr 26, 2008 #10
    Look into the proof of Coshi theory - Riman

    f'(z)=f'(x)=f'(y)


    no f'(z)=f'(x)+f'(y)
     
  12. Apr 26, 2008 #11

    Dick

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    What have derivatives got to do with it? You were posting limit problems. Or something. I give up. I don't know what your question or problem is.
     
  13. Apr 26, 2008 #12
    I support the second solution
    And I am not sure
     
  14. Apr 26, 2008 #13

    Dick

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    If you are asking does |z|^2/z approach 0 as z approaches 0. Then, it does. You're first argument is fine. If you think there is a problem because the limit of z* as z approaches 0 is path dependent, then you are wrong. The limit is 0.
     
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