# Help on epsilon delta proof of discontinuity

1. Jul 13, 2011

### dillingertaco

1. The problem statement, all variables and given/known data
Prove the function f(x)= { 4 if x=0; x^2 otherwise
is discontinuous at 0 using epsilon delta.

2. Relevant equations
definiton of discontinuity in this case:
there exists an e>0 such that for all d>0 if |x-0|<d, |x^2-4|>e

3. The attempt at a solution
I'm confused because if we include ALL delta >0, eventually (namely, x around +/- 2) |x^2-4| will be less than e for all e>0 which seems to me to point to it being continuous at 0 when it clearly is NOT. Is there something built into the definition that ignores large values of delta which makes the interval around x too large?

So my way:
Assume it is continuous at 0. Let e=1
Then |x^-4|<1 when |x|<d for some d.

From here I want to say d<1 and so |x^2-4|>2 for all |x|<d which would be the contradiction but I don't think that's how I formally say it...

2. Jul 13, 2011

### micromass

Staff Emeritus
Hi dillingertaco!

Your definition of discontinuity is wrong. Continuity in 0 says that

"For all $\varepsilon >0$, there is a $\delta>0$, such that for all x holds that $|x-0|<\delta~\Rightarrow~|x^2-4|<\varepsilon$"

So, the converse of this statement is

"There is a $\varepsilon >0$, such that for all $\delta >0$, there is an x such that $|x-0|<\delta$ but $|x^2-4|\geq \varepsilon$"

So you don't need things to hold for all x such that $|x-0|<\delta$, but only for one specified x. Does that clear things up?

3. Jul 13, 2011

### dillingertaco

Hello micromass, I appreciate the response!

I'm getting confused here how to formalize this. So for all $\delta$, we can find such an x as you described. It makes sense in the terms of $delta = 1$, look at x= 1/2, with $\epsilon = 1$.
I see that this case works, but how do do this for every delta? I can say now that I found an x for any $\delta\geq 1$ but now what about $delta\leq 1$ without going through the same argument?

I guess we could let x= 1/n and say there exists an n by the archimedean property that this is true and therefore we can do it that way, but it seems to me I used to do this without using this step... but maybe I didn't?

4. Jul 13, 2011

### micromass

Staff Emeritus
So you take $\varepsilon=1$. For all $\delta>0$, you must find an x such that

$$|x^2-4|\geq 1,~\text{but}~|x|<\delta$$

Indeed, if $\delta >1$, you can take x=1.
And if $\delta\leq 1$, what if you take $x=\delta/2$?

5. Jul 13, 2011

### dillingertaco

That's a better idea. Then $|\frac{\delta}{2}^{2}-4|\geq 1$ since $\delta/2\leq1/2$

Would there be a way to do this without the caveat when $\delta>1$. it seems very inelegant.

6. Jul 13, 2011

### micromass

Staff Emeritus
Epsilon-delta stuff always tend to be inelegant But no, I don't think there is an easier/more elegant way of doing this.

7. Jul 13, 2011

### dillingertaco

I think the way I'm thinking of it defining x=min(d/2,1). That would make me happy.

Thank you so much for your help.

8. Jul 13, 2011

### micromass

Staff Emeritus
Oh, yes, if you find that more elegant, then you can always do that of course