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

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- Dec 1st 2010, 06:37 AMalexandrosproving continuity
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 - Dec 1st 2010, 06:52 AMchisigma
The limit...

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Dec 1st 2010, 07:19 AMalexandros
- Dec 1st 2010, 12:21 PMchisigma
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

$\displaystyle \chi$ $\displaystyle \sigma$ - Dec 1st 2010, 02:09 PMalexandros
How,then do we otherwise

**prove**that the function is discontinuous at (0,0) - Dec 1st 2010, 02:35 PMnerdo
- Dec 1st 2010, 02:37 PMPlato
- Dec 1st 2010, 02:40 PMnerdo
- Dec 1st 2010, 02:42 PMPlato
- Dec 1st 2010, 02:52 PMnerdo
I think personally if one used the theorem below, the question is trivial:

Attachment 19924 - Dec 1st 2010, 02:55 PMPlato
- Dec 1st 2010, 02:59 PMnerdo
Oh ok i see what you mean, i guess you are right.

- Dec 1st 2010, 03:32 PMalexandros
- Dec 1st 2010, 04:12 PMalexandros
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** - Dec 2nd 2010, 11:29 AMSammySproving continuity

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.