1. Originally Posted by SammyS

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).

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??

2. Originally Posted by SammyS

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).

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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