Help on epsilon delta proof of discontinuity

[dillingertaco]

Homework Statement

Prove the function f(x)= { 4 if x=0; x^2 otherwise
is discontinuous at 0 using epsilon delta.

Homework Equations

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

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...

 

[micromass]

Hi dillingertaco!

Homework Statement

Prove the function f(x)= { 4 if x=0; x^2 otherwise
is discontinuous at 0 using epsilon delta.

Homework Equations

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

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

"For all \varepsilon &gt;0, there is a \delta&gt;0, such that for all x holds that |x-0|&lt;\delta~\Rightarrow~|x^2-4|&lt;\varepsilon"

So, the converse of this statement is

"There is a \varepsilon &gt;0, such that for all \delta &gt;0, there is an x such that |x-0|&lt;\delta but |x^2-4|\geq \varepsilon"

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

 

[dillingertaco]

Hello micromass, I appreciate the response!

So you don't need things to hold for all x such that |x-0|&lt;\delta, but only for one specified x.

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?

 

[micromass]

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?

So you take \varepsilon=1. For all \delta&gt;0, you must find an x such that

|x^2-4|\geq 1,~\text{but}~|x|&lt;\delta

Indeed, if \delta &gt;1, you can take x=1.
And if \delta\leq 1, what if you take x=\delta/2?

 

[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&gt;1. it seems very inelegant.

 

[micromass]

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&gt;1. it seems very inelegant.

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

 

[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.

 

[micromass]

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.

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