Originally Posted by

**SammyS**

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

The problem with $\displaystyle f(x,y)$ is that it is undefined along the line $\displaystyle y=-x$, except at (0,0). Let's make a sequence $\displaystyle (x_n, y_n)$ which approaches (0,0) along a path that approaches $\displaystyle y=-x$. One such path is $\displaystyle y=-x+ax^2$. Define $\displaystyle \displaystyle x_n = {1\over n}$, then $\displaystyle \displaystyle y_n=-{1\over n}+{a\over {n^2}}$, so $\displaystyle \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 $\displaystyle f(0,0)=0$, we have that $\displaystyle f$ is not continuous at (0,0).

**Now for your ε-δ proof:**

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

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

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

Fill in the rest of the details.