proving continuity

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• Dec 1st 2010, 06:37 AM
alexandros
proving 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 AM
chisigma
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 AM
alexandros
Quote:

Originally Posted by chi sigma
The limit...

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

Thank you, but if you read my post i asked for a proof of that discontinuity by using ε-δ definition for discontinuity
• Dec 1st 2010, 12:21 PM
chisigma
Quote:

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 '$\displaystyle \varepsilon-\delta$ 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 PM
alexandros
How,then do we otherwise prove that the function is discontinuous at (0,0)
• Dec 1st 2010, 02:35 PM
nerdo
Quote:

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)
• Dec 1st 2010, 02:37 PM
Plato
Quote:

Originally Posted by alexandros
How,then do we otherwise prove that the function is discontinuous at (0,0)

Can you show that $\displaystyle \displaystyle\lim _{(x,y) \to (0,0)} \frac{{xy}}{{x + y}}$ does not exist?

In order for $\displaystyle f(x,y)$ to be continuous at $\displaystyle (0,0)$ that limit must exist.
• Dec 1st 2010, 02:40 PM
nerdo
Quote:

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
• Dec 1st 2010, 02:42 PM
Plato
Quote:

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?
• Dec 1st 2010, 02:52 PM
nerdo
Quote:

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:

Attachment 19924
• Dec 1st 2010, 02:55 PM
Plato
Quote:

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.
• Dec 1st 2010, 02:59 PM
nerdo
Oh ok i see what you mean, i guess you are right.
• Dec 1st 2010, 03:32 PM
alexandros
Quote:

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 ??
• Dec 1st 2010, 04:12 PM
alexandros
Quote:

Originally Posted by Plato
Can you show that $\displaystyle \displaystyle\lim _{(x,y) \to (0,0)} \frac{{xy}}{{x + y}}$ does not exist?

In order for $\displaystyle f(x,y)$ to be continuous at $\displaystyle (0,0)$ that limit must exist.

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 AM
SammyS
proving continuity
Quote:

Originally Posted by alexandros
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 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 $\displaystyle f$ 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).

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