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Math Help - proving continuity

  1. #1
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    proving continuity

    given the function:

    f(x,y)= \frac{xy}{x+y} if (x,y)\neq 0

    f(x,y) =0 if (x,y)=(0,0)

    prove whether it is continuous or discintinuous at (0,0) by using the ε-δ definition for continuity
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    MHF Contributor chisigma's Avatar
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    The limit...

    \displaystyle \lim_{(x,y) \rightarrow (0,0)} \frac{x y}{x+y}

    ... doesn't exist so that the function is discontinous in (0,0)...

    Kind regards

    \chi \sigma
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    Quote Originally Posted by chi sigma View Post
    The limit...

    \displaystyle \lim_{(x,y) \rightarrow (0,0)} \frac{x y}{x+y}

    ... doesn't exist so that the function is discontinuous in (0,0)...

    Kind regards

    \chi \sigma
    Thank you, but if you read my post i asked for a proof of that discontinuity by using ε-δ definition for discontinuity
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    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by alexandros View Post
    Thank you, but if you read my post i asked for a proof of that discontinuity by using ε-δ definition for discontinuity
    The problem is that the so called ' \varepsilon-\delta proof' is adequate to demostrate the continuity and not the discontinuity of a function... in this particular case the limit...

    \displaystyle \lim_{(x,y) \rightarrow (0,0)} \frac{x y}{x+y} (1)

    ... is 'almost always' 0... almost because when is x+y=0 the (1) is not defined...

    Kind regards

    \chi \sigma
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    How,then do we otherwise prove that the function is discontinuous at (0,0)
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    Quote Originally Posted by alexandros View Post
    How,then do we otherwise prove that the function is discontinuous at (0,0)
    Here is away:

    Choose (Xn,Yn)=(1/n,1/n) then (Xn,Yn)-> (0,0) but F(Xn,Yn)=1/2 so the function is discontinuous at (0,0)
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    Quote Originally Posted by alexandros View Post
    How,then do we otherwise prove that the function is discontinuous at (0,0)
    Can you show that \displaystyle\lim _{(x,y) \to (0,0)} \frac{{xy}}{{x + y}} does not exist?

    In order for f(x,y) to be continuous at (0,0) that limit must exist.
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    Quote Originally Posted by nerdo View Post
    Here is away:

    Choose (Xn,Yn)=(1/n,1/n) then (Xn,Yn)-> (0,0) but F(Xn,Yn)=1/2 so the function is discontinuous at (0,0)
    I think this is by far the best way to show that this discontinuous
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  9. #9
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    Quote Originally Posted by nerdo View Post
    I think this is by far the best way to show that this discontinuous
    Why do you this there is such a notion as there is best way?
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  10. #10
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    Quote Originally Posted by Plato View Post
    Why do you this there is such a notion as there is best way?
    I think personally if one used the theorem below, the question is trivial:


    proving continuity-untitled.jpg
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    Quote Originally Posted by nerdo View Post
    I think personally if one used the theorem below, the question is trivial:
    But using a theorem does not mean the learner understands to basic concept.
    A proof is a proof is a proof.
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  12. #12
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    Oh ok i see what you mean, i guess you are right.
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  13. #13
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    Quote Originally Posted by nerdo View Post
    Here is away:

    Choose (Xn,Yn)=(1/n,1/n) then (Xn,Yn)-> (0,0) but F(Xn,Yn)=1/2 so the function is discontinuous at (0,0)
    But F(Xn,Yn)= 1/2n and not 1/2 ??
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    Quote Originally Posted by Plato View Post
    Can you show that \displaystyle\lim _{(x,y) \to (0,0)} \frac{{xy}}{{x + y}} does not exist?

    In order for f(x,y) to be continuous at (0,0) that limit must exist.
    To show that : \displaystyle\lim _{(x,y) \to (0,0)} \frac{{xy}}{{x + y}} does not exist?

    We must show that: for all real Nos m there exists an ε>0 such for all δ>0 there exists a pair (x,y) such that : |x|<δ and |y|<δ and |\frac{xy}{x+y}-m|\geq\epsilon

    However to show discontinuity at (0,0) we must show that:

    There exists ε>0 such that for all δ>0 there exists (x,y) such that:|x|<δ and |y|<δ and |\frac{xy}{x+y}|\geq\epsilon

    Two comletely different things
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  15. #15
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    proving continuity

    Quote Originally Posted by alexandros View Post
    given the function:

    f(x,y)= \frac{xy}{x+y} if (x,y)\neq 0

    f(x,y) =0 if (x,y)=(0,0)

    prove whether it is continuous or discontinuous at (0,0) by using the ε-δ definition for continuity

    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.

    Last edited by SammyS; December 2nd 2010 at 01:23 PM.
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