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
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.
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.
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)
Probably total rubbish.
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)).Anyway my attempt. x<0, abs(x)= -x so -x< sigma if sigma =min(1,e)
The case x >= 0 is very similar but simpler since |x| = x.
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.I don't really understand what the definition intuitively means.
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 . 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.