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Finding a Limit in two variables

  1. Jun 25, 2012 #1
    1. The problem statement, all variables and given/known data
    f(1/u,0)=1 and f(0,1/u)=-1 for all positive (integer) values of u. Prove whether or not the limit as (x,y) ->(0,0) exists.

    2. Relevant equations

    none.

    3. The attempt at a solution

    I argue that 0,0 is not in the domain of the function, but this neglects the behavior of f(x,y). So I feel like I'm missing a clue. It seems similar to the binary function: f(x)=-1 when x>0 and f(x)=-1 when x<0. But I'm unsure how to solve and prove the limit.

    Edit:[ok, I'm saying that if (1/u) were to equal 0, then n would end up being infinity (not a positive integer), but again, this doesn't account for the -1/1 values of f(x,y)]
     
    Last edited: Jun 25, 2012
  2. jcsd
  3. Jun 25, 2012 #2
    By definition of limit, x and y would never be equal to zero.
     
  4. Jun 25, 2012 #3
    Why is that? Can you not take: {Lim as u→∞[f(1/u,0)]}=f(0,0)

    I was thinking that if I evaluate: [itex]Lim_{u\rightarrow\infty}[/itex]f(1/u,0)[itex]\neq[/itex][itex]Lim_{u\rightarrow\infty}[/itex]f(0,1/u)
    Because 1≠-1

    Any help is appreciated
     
  5. Jun 25, 2012 #4

    LCKurtz

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    You could only claim that if f(0,0) was defined, which it isn't, and if it were and it was actually equal to that limit.

    You are getting close. Remember that for a two variable function, if$$
    \lim_{(x,y)\rightarrow (a,b)} f(x,y)= L$$ that requires that you get ##L## as ##(x,y)\rightarrow (a,b)## along any path. What can you conclude from that?
     
  6. Jun 25, 2012 #5
    I conclude that because a limit for a point must be equivalent from any real path, [itex]\lim_{(x,y)\rightarrow (a,b)} f(x,y)= 1[/itex] via path [itex]f(x,y)=(1/u,0)[/itex] but [itex]\lim_{(x,y)\rightarrow (a,b)} f(x,y)= -1\neq 1[/itex] via path [itex]f(x,y)=(0,1/u)[/itex].

    Is such a conclusion sound ?
     
  7. Jun 25, 2012 #6

    LCKurtz

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    Your original question was whether or not the limit exists. What do you conclude about that?
     
  8. Jun 25, 2012 #7
    I conclude that because a limit for a point must be equivalent from any real path and [itex]\lim_{(x,y)\rightarrow (a,b)} f(x,y)= 1[/itex] via path [itex]f(x,y)=(1/u,0)[/itex] but [itex]\lim_{(x,y)\rightarrow (a,b)} f(x,y)= -1\neq 1[/itex] via path [itex]f(x,y)=(0,1/u)[/itex], the limit at (x,y)=(0,0) does not exist.
     
  9. Jun 25, 2012 #8
    I should have said x OR y are non zero.
    That is, by definition of limit, (x,y)≠(0,0)
     
  10. Jun 25, 2012 #9

    LCKurtz

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    That's right.
     
  11. Jun 25, 2012 #10
    Thanks for the help !
     
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