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Proving that a limit does not exist

  1. Jan 16, 2014 #1
    1. The problem statement, all variables and given/known data

    See attached pic

    2. Relevant equations

    See attached pic

    3. The attempt at a solution

    See attached pic

    Basically I need to know if the way I've done this problem is OK. I approached the point (0,0) along path y=x2, substituted 0 for x and got a value of 1. Fine

    Then I approached along the Y axis, meaning x is always 0. Because x is always 0 and the numerator, the value of the function will be zero for all y.

    Is this an OK argument?

    Thanks
     

    Attached Files:

  2. jcsd
  3. Jan 16, 2014 #2

    haruspex

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    Your argument is sound. (But why not just approach along y=x?)
     
  4. Jan 16, 2014 #3
    Because theres an x-y on the bottom, therefore this will give a 0 on the denominator for all x
     
  5. Jan 16, 2014 #4

    haruspex

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    Right. So? Doesn't that prove there's no such limit?
     
  6. Jan 17, 2014 #5
    So are you saying if the limit is not continuous in a neighbourhood of (x0,y0) then that's sufficient proof that the limit does not exist?
     
  7. Jan 17, 2014 #6

    haruspex

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    I'm saying that if you have an infinite sequence of points converging to (x0, y0) such that the function is not defined at any of them then it cannot be continuous at (x0, y0).
     
  8. Jan 17, 2014 #7
    Ok, I think that's what I meant. I have another question which I've been stuck on for hours: How do I find the derivative with respect to x of

    xsin(x)

    I'm really stuck on this, been looking for a good explaination on how to do it. I've a calculus exam tomorrow :cry:
     
  9. Jan 17, 2014 #8

    Dick

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    ##x=e^{log(x)}##. So ##x^{sin(x)}=e^{log(x) sin(x)}##. Now use the chain rule on that. In the future, I'd open a new thread for unrelated questions.
     
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