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Two-Path Test:Fail?

  1. Aug 9, 2012 #1
    Well, I've been calculating some limits of several variable functions and got confused with something: everybody knows that if taking a path the limit depends on the path chosen then the limit doesn't exist. But, if you consider the limit

    lim (x,y) -> (0,0) [itex]\frac{y^{2}x}{x^{2}+y^{2}}[/itex]

    that can be easily calculated( function that converges to 0 multiplied by limited function ). ALthough, if you choose the path

    y=[itex]\sqrt{\frac{c}{x-c}}x[/itex]

    you'll end up with a limit equals to c, in other words, it depends oof the path you choose what would be enough for us to say the limit doesn't exist. The problem is we know it exists and my question is: did i forget to consider any important hypothesis or condition? How can the path test be applied correctly then?
    I appreciate your attention.
     
  2. jcsd
  3. Aug 9, 2012 #2


    The path must be chosen as to make sense when the variable(s) approach the wanted point. When [itex]\,x\to 0\,[/itex] , the expression
    [tex]\sqrt{\frac{c}{x-c}}\,[/tex]
    becomes undefined (over the reals) as the fraction into the square root becomes negative, no matter what number [itex]\,c\neq 0\,[/itex] is.

    DonAntonio
     
  4. Aug 9, 2012 #3
    Thanks DonAntonio.
    It's so obvious yet i didn't noticed it.
     
  5. Aug 9, 2012 #4

    Don't worry about it (it occurs to us all, no matter whether you're a beginning undergraduate or a graduate one.

    Just learn this lesson and try to remeber it in the future...where new mistakes await for us all, too. :)

    DonAntonio
     
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