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Math Help - How to prove that this limit does exists...

  1. #1
    Newbie HypatiaDaVinci's Avatar
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    How to prove that this limit does exists...

    How can I correctly justify, demonstrate, and properly determine that this limit exists?

    lim (x,y) --> (0,0) \sqrt{x^2+y^2}
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  2. #2
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    Quote Originally Posted by HypatiaDaVinci View Post
    How can I correctly justify, demonstrate, and properly determine that this limit exists?
    lim (x,y) --> (0,0) \sqrt{x^2+y^2}
    The distance .

    So
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  3. #3
    Newbie HypatiaDaVinci's Avatar
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    Thank you very much for the answer. How do you get to the conclusion that \varepsilon=\delta I mean... how do you directly relate them?
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  4. #4
    Super Member TheChaz's Avatar
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    Quote Originally Posted by HypatiaDaVinci View Post
    Thank you very much for the answer. How do you get to the conclusion that \varepsilon=\delta I mean... how do you directly relate them?
    To prove this by the definition, you should state that
    "For all x,y in R, for all epsilon > 0, there exists a delta > 0 such that [d(x, y) < delta] implies [sqrt(x^2 + y^2) < epsilon]
    since "d(x, y)" and "sqrt(x^2 + y^2)" are the same thing, you can choose delta to be the epsilon that is given.
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  5. #5
    Newbie HypatiaDaVinci's Avatar
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    Thank you very much, I think I got it now
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