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How to prove that a limit of a two-variable function does not exist?

  1. Mar 27, 2012 #1
    9065he.jpg


    I need to prove this limit does NOT exist. I know that the denominator cancels out, however my lecturer says the limit does not exist because we cannot find a circle with radius [itex]\delta[/itex] that contains the point (0,0), as the function is not defined for all y=x. Here is a little drawing of the domain.

    Is my lecturer's explanation valid?


    Thanks in advance,
    Michael
     
  2. jcsd
  3. Mar 27, 2012 #2
    Rationalize it
     
  4. Mar 27, 2012 #3
    I guess his explanation makes sense, but that's a funny way of putting it.
     
  5. Mar 27, 2012 #4
    Can you think of another way? If you do, please post
     
  6. Mar 27, 2012 #5

    SammyS

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    Well, that limit certainly does exist --- at least it exists using the mufti-variable definition of limit with which I am familiar.

    [itex]\displaystyle \text{If }x\ne y\,,\text{ then }\frac{x^2-y^2}{x-y}=\frac{(x-y)(x+y)}{x-y}=x+y\,.[/itex] No matter what continuous path you take (in the domain of the given function) to approach (0,0), x+y → 0 .

    On the other hand, the function is not continuous at (x,y) = (0,0) because it's not defined at (x,y) = (0,0) .
     
  7. Mar 27, 2012 #6

    LCKurtz

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    I think it is common for functions of two variables to be required to be defined in a punctured disk around (a,b) in order to claim a limit as ##(x,y)\to (a,b)## of the function exist. That is presumably required in this case, given the lecturer's statement.
     
  8. Mar 28, 2012 #7
    Well, as far as I know, the limit does not have to be defined at the point itself (0,0), but there must exist at least one disc that contains it.
     
  9. Mar 28, 2012 #8

    LCKurtz

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    Yes, there must be a punctured disk around (a,b) in the domain of the function, which you don't have.
     
  10. Mar 28, 2012 #9
    Alright, thanks a lot!
     
  11. Mar 28, 2012 #10

    HallsofIvy

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    Even the path y= x??

     
  12. Mar 28, 2012 #11

    SammyS

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    y=x is not in the domain of [itex]\displaystyle \frac{x^2-y^2}{x-y}\ .[/itex]

    It appears that Calculus textbooks don't agree on the definition of the limit for functions of more than one variable.
     
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