# Help on epsilon delta proof of discontinuity

• dillingertaco
Glad I could help!In summary, the conversation discusses proving the function f(x)= { 4 if x=0; x^2 otherwise is discontinuous at 0 using epsilon-delta. The definition of discontinuity is defined as the existence of an epsilon greater than 0, such that for all delta greater than 0, if |x-0|<delta, then |x^2-4|>epsilon. The attempted solution is discussed and it is discovered that the definition of discontinuity used is incorrect. The correct definition states that for all epsilon greater than 0, there is a delta greater than 0, such that for all x, if |x-0|<delta, then |x^2
dillingertaco

## Homework Statement

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

## Homework 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

## 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...

Hi dillingertaco!

dillingertaco said:

## Homework Statement

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

## Homework 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

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?

Hello micromass, I appreciate the response!

micromass said:
So you don't need things to hold for all x such that $|x-0|<\delta$, 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?

dillingertaco said:
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 $\delta>0$, 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 $x=\delta/2$?

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.

dillingertaco said:
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.

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

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.

dillingertaco said:
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

## 1. What is an epsilon delta proof?

An epsilon delta proof is a mathematical method used to formally prove the existence of a limit of a function. It involves choosing a small positive number (epsilon) and showing that for any input value (delta), the output value of the function is within epsilon distance of the limit.

## 2. Why is an epsilon delta proof used to prove discontinuity?

An epsilon delta proof is used to prove discontinuity because it allows for a rigorous and precise analysis of the behavior of a function at a specific point. By showing that a function does not have a limit at a certain point, we can prove that it is discontinuous at that point.

## 3. How do I construct an epsilon delta proof for discontinuity?

To construct an epsilon delta proof for discontinuity, you first need to define the function and the point at which you want to prove discontinuity. Then, choose a small value for epsilon and show that for any delta within that range, there exists an input value for which the output value is outside the epsilon range.

## 4. Can an epsilon delta proof be used for all types of discontinuity?

Yes, an epsilon delta proof can be used for all types of discontinuity, including removable, jump, and infinite discontinuities. However, the specific approach and method may vary depending on the type of discontinuity.

## 5. Are there any limitations to using an epsilon delta proof for discontinuity?

While an epsilon delta proof is a powerful tool for proving discontinuity, it does have its limitations. It may not work for extremely complex or undefined functions, and it also relies on the assumption that the limit of the function exists. Additionally, some discontinuities may be difficult to prove using this method, requiring alternative approaches.

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