1. ## Proving continuity

Use the given definition of continuity to prove f is continuos at 0.

Definition: For all £>0, there exists a k such that abs(f(x)-f(a))<£ for all x such at
abs(x-a)<k

F(x) = 1+x^2 when x>_0 , f(x)= 1-x^3 when x<0

I have trouble understanding the implications of the definition of continuity.

I have that abs(f(x)-f(0))=abs(f(x)-1) but don't what to do next.

2. Originally Posted by poirot
Use the given definition of continuity to prove f is continuos at 0.
Definition: For all £>0, there exists a k such that abs(f(x)-f(a))<£ for all x such at
abs(x-a)<k
F(x) = 1+x^2 when x>_0 , f(x)= 1-x^3 when x<0
I have trouble understanding the implications of the definition of continuity.
I have that abs(f(x)-f(0))=abs(f(x)-1) but don't what to do next.
You have to consider two cases:
$\left| {F(x) - F(0)} \right| = \left\{ {\begin{array}{rl} {x^2 ,} & {x > 0} \\ {-x^3 ,} & {x < 0} \\ \end{array} } \right.$.

3. Can you show me how to do it?

4. Originally Posted by poirot
Can you show me how to do it?
Of course I could. But it is your problem.
You have shown no effort on your part.
Now what do you know about the $\epsilon/\delta$ method of proof?

5. Nothing, I haven't encountered it before.

6. Originally Posted by poirot
Nothing, I haven't encountered it before.
How can you say with any sincerity?
Originally Posted by poirot
Use the given definition of continuity to prove f is continuos at 0. Definition: For all £>0, there exists a k such that abs(f(x)-f(a))<£ for all x such at
abs(x-a)<k
F(x) = 1+x^2 when x>_0 , f(x)= 1-x^3 when x<0
That is what this entire question is about. You posted it.
Did you post a question that you know absolutely nothing about?
If so, that is exceedingly odd.

7. I know the definition but I don't really get it and don't know how to apply it to questions. I meant I have not done a question like this from first principles. So if you could post half of the method I will probably get it. I would prefer if you posted it all but you seem averse to that.

8. Originally Posted by poirot
I know the definition but I don't really get it and don't know how to apply it to questions. I meant I have not done a question like this from first principles. So if you could post half of the method I will probably get it. I would prefer if you posted it all but you seem averse to that.
If that is the truth, the show us.
Suppose $\varepsilon > 0$ then let $\delta = \min \left\{ {1,\varepsilon } \right\}$.
Now YOU show that $\left| {x - 0} \right| < \delta \, \Rightarrow \,\left| {F(x) - 1} \right| < \varepsilon .$

9. I don't really understand what the definition intuitively means. An explanation would be welcomed.

In post 2,how have got that equation?

Anyway my attempt. x<0, abs(x)= -x so -x< sigma if sigma =min(1,e)

abs(f(x)-1)=abs(x^3)= (-x)^3<sigma^3<e.

Probably total rubbish.

10. Originally Posted by poirot
I don't really understand what the definition intuitively means. An explanation would be welcomed.
In post 2,how have got that equation?
Anyway my attempt. x<0, abs(x)= -x so -x< sigma if sigma =min(1,e)
abs(f(x)-1)=abs(x^3)= (-x)^3<sigma^3<e.
Probably total rubbish.
Yes, it is rubbish.
Over and out

11. Anyway my attempt. x<0, abs(x)= -x so -x< sigma if sigma =min(1,e)

abs(f(x)-1)=abs(x^3)= (-x)^3<sigma^3<e.
I think this is pretty correct. I would expand it a little as follows. Let δ (it is called delta) be min(1,ε) and suppose that 0 < -x < δ. Then |f(x) - 1| = |x^3| = (-x)^3 < δ^3 <= δ (since δ <= 1) <= ε (since δ = min(1,ε)).

The case x >= 0 is very similar but simpler since |x| = x.

I don't really understand what the definition intuitively means.
Looks like you need a tutorial on epsilon-delta. I don't have good links, but this one I found just now seems all right. Look especially at this intuitive explanation using Flash.

12. eminently more helpful. How did you know to set delta to be min(1,epsilon)?

13. In this example, this was Plato's suggestion. The intuition is that we must make x^2 to be < ε when x < δ. Since δ^2 < δ when 0 < δ < 1, it is sufficient to choose such δ that both δ < 1 and δ < ε hold. It is also possible to choose $\delta = \min(\sqrt{\varepsilon},\sqrt[3]{\varepsilon})$. And of course, any smaller value of δ also works, such as min(1,ε) / 2.

I general, there is no algorithm for choosing δ given ε; it completely depends on the function. The definition of the continuity just requires that for every ε some δ exists. When proving continuity from the definition, one can use any means to find a δ. One becomes better at this with practice. It is even possible to prove the existence of δ by contradiction instead of coming up with a specific example, i.e., to show that the assumption that no suitable δ exists is absurd.