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Math Help - proving continuity

  1. #16
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    Quote Originally Posted by SammyS View Post

    I realize that you need an ε-δ
    proof, but first let's find a sequence that shows that f is discontinuous at (0,0). This may help with your ε-δ proof.

    The problem with
    f(x,y) is that it is undefined along the line y=-x, except at (0,0). Let's make a sequence (x_n, y_n) which approaches (0,0) along a path that approaches y=-x. One such path is y=-x+ax^2. Define \displaystyle x_n = {1\over n}, then \displaystyle y_n=-{1\over n}+{a\over {n^2}}, so \displaystyle f(x_n,y_n)= {{\left({1\over n}\right)\left(-{1\over n}+{a\over {n^2}}\right)} \over{\left({1\over n}\right)+\left(-{1\over n}+{a\over {n^2}}\right)}}={{-{1\over {n^2}}+{a\over {n^3}}}  \over{a\over {n^2}}} = -{1\over a}+ {1\over n}\to -{1\over a} \text{ as } n \to \infty .

    Since
    f(0,0)=0, we have that f is not continuous at (0,0).

    Now for your ε-δ proof:

    Rather than showing that : \displaystyle\lim _{(x,y) \to (0,0)} \frac{{xy}}{{x + y}} does not exist, just show that it's not 0. Modify your statement to have m=0.

    Given any \delta>0, let x={\delta\over2} (may have to tweak this a little). Let y=-x-x^2=-{\delta\over2}-{{\delta^2}\over{4}}.

    Then \displaystyle f(x,y)= {{\left({\delta\over 2}\right)\left(-{\delta\over2}-{{\delta^2}\over{4}}\right)} \over{\left({\delta\over 2}\right)+\left(-{\delta\over2}-{{\delta^2}\over{4}}\right)}}=1+{\delta\over 2}

    Fill in the rest of the details.

    Thank you

    But in the ε-δ proof I cannot find the ε

    Can you help a little more??
    Last edited by alexandros; December 2nd 2010 at 04:57 PM. Reason: correction
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  2. #17
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    Quote Originally Posted by SammyS View Post

    I realize that you need an ε-δ
    proof, but first let's find a sequence that shows that f is discontinuous at (0,0). This may help with your ε-δ proof.

    The problem with
    f(x,y) is that it is undefined along the line y=-x, except at (0,0). Let's make a sequence (x_n, y_n) which approaches (0,0) along a path that approaches y=-x. One such path is y=-x+ax^2. Define \displaystyle x_n = {1\over n}, then \displaystyle y_n=-{1\over n}+{a\over {n^2}}, so \displaystyle f(x_n,y_n)= {{\left({1\over n}\right)\left(-{1\over n}+{a\over {n^2}}\right)} \over{\left({1\over n}\right)+\left(-{1\over n}+{a\over {n^2}}\right)}}={{-{1\over {n^2}}+{a\over {n^3}}}  \over{a\over {n^2}}} = -{1\over a}+ {1\over n}\to -{1\over a} \text{ as } n \to \infty .

    Since
    f(0,0)=0, we have that f is not continuous at (0,0).

    Now for your ε-δ proof:

    Rather than showing that : \displaystyle\lim _{(x,y) \to (0,0)} \frac{{xy}}{{x + y}} does not exist, just show that it's not 0. Modify your statement to have m=0.

    Given any \delta>0, let x={\delta\over2} (may have to tweak this a little). Let  y=-x-x^2=-{\delta\over2}-{{\delta^2}\over{4}}.

    Then \displaystyle f(x,y)= {{\left({\delta\over 2}\right)\left(-{\delta\over2}-{{\delta^2}\over{4}}\right)} \over{\left({\delta\over 2}\right)+\left(-{\delta\over2}-{{\delta^2}\over{4}}\right)}}=1+{\delta\over 2}

    Fill in the rest of the details.

    Another thing to notice here is that,if we put :y =  y=-x-x^2=-{\delta\over2}-{{\delta^2}\over{4}},since we want |y|<δ, that substitution will lead us to the inequality: δ(δ-2)<0

    Αnd since δ>0 we must have δ>2 ,thus imposing a restriction on δ>0

    But we must show that for all δ>0 and not just for those greater than 2 the inequalities : |x|<δ ,|y|<δ |\frac{xy}{x+y}|\geq\epsilon must hold ,for a certain ε>0
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