given the function:
f(x,y)= $\displaystyle \frac{xy}{x+y}$ if $\displaystyle (x,y)\neq 0$
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
The problem is that the so called '$\displaystyle \varepsilon-\delta$ proof' is adequate to demostrate the continuity and not the discontinuity of a function... in this particular case the limit...
$\displaystyle \displaystyle \lim_{(x,y) \rightarrow (0,0)} \frac{x y}{x+y}$ (1)
... is 'almost always' 0... almost because when is $\displaystyle x+y=0$ the (1) is not defined...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
To show that :$\displaystyle \displaystyle\lim _{(x,y) \to (0,0)} \frac{{xy}}{{x + y}}$ 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 $\displaystyle |\frac{xy}{x+y}-m|\geq\epsilon$
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 $\displaystyle |\frac{xy}{x+y}|\geq\epsilon$
Two comletely different things
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.