1. ## 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

2. 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

$\chi$ $\sigma$

3. Originally Posted by chi sigma
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

4. Originally Posted by alexandros
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

$\chi$ $\sigma$

5. How,then do we otherwise prove that the function is discontinuous at (0,0)

6. Originally Posted by alexandros
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)

7. Originally Posted by alexandros
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.

8. Originally Posted by nerdo
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

9. Originally Posted by nerdo
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?

10. Originally Posted by Plato
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:

11. Originally Posted by nerdo
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.

12. Oh ok i see what you mean, i guess you are right.

13. Originally Posted by nerdo
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 ??

14. Originally Posted by Plato
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

15. ## proving continuity

Originally Posted by alexandros
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).

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