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Stuck Proving a Limit Doesn't exist

  1. Feb 18, 2010 #1
    1. The problem statement, all variables and given/known data
    The problem is to either find the limit or show that it does not exist [tex]lim_{(x,y)\rightarrow(2,-2)}\frac{4-xy}{4+xy}[/tex]

    I've been able to do similar problems to this such as
    [tex]lim_{(x,y)\rightarrow(0,0)}\frac{xy}{x^2+y^2}[/tex] where I took two different paths to the limit and found that they were not equal and so it didn't exist. However, for this one I can't seem to pick a function that gives me a limit that exists let alone two functions that give me two different limits.

    I've tried coming from the following paths for the problem
    [tex]y=-x[/tex]
    [tex]y=x-4[/tex]
    [tex]y=-2[/tex]
    [tex]y=x^2-6[/tex]
    [tex]x=2[/tex]

    No matter which one I do I can't seem to get anything to cancel out in order to simplify it to one that I can perform the limit on. Are there two functions that I can use to get this limit and how would I find these. If there aren't then how would I prove that this limit exists?
     
  2. jcsd
  3. Feb 18, 2010 #2

    Dick

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    I don't see anything wrong with y=(-x). Put that in and let x->2. What do you conclude?
     
  4. Feb 18, 2010 #3
    When I plug in y=-x the limit becomes
    [tex]lim_{(x,y)\rightarrow(2,-2)}\frac{4+x^2}{4-x^2}[/tex]
    But then when I substitute in the values I still end up with 8/0 which is undefined. Am I doing something wrong when I plug in that function?
     
  5. Feb 18, 2010 #4

    Dick

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    You don't really just 'plug in'. You think about what happens as x approaches 2. And you are right, the numerator goes to 8 and the denominator goes to 0. I would say the limit doesn't exist. The quotient just gets larger and larger as x->2.
     
  6. Feb 18, 2010 #5
    So I don't have to show that multiple paths equal different things in this case just that one of the paths doesn't exist?
     
  7. Feb 18, 2010 #6

    Dick

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    Right. In this case there is no limit along ANY path. No matter what the path you still get an 8/0 type quotient.
     
  8. Feb 18, 2010 #7
    Thank you. That makes so much sense now. I was under the impression that we had to find two functions with different values but now I can see that this works too.
     
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