Epsilon-Delta proofs, once again

[dzogi]

I'm trying to understand \epsilon-\delta proofs, but I'm having some trouble. For example, if we want to prove that \lim_{x\rightarrow2}x^3=8, starting from |x^3-8| we get to something like

|x-2||x^2+2x+4|

And this is what confuses me: we conjecture that |x-2|<1, then |x|<3, so we get

|x-2||x^2+2x+4| \leq |x-2|||x|^2+2|x|+4|<|x-2||3^2+2*3+4|=19|x-2| then we can easily find out what \delta is. But, why do we make that assumption when we're working with every \epsilon>0? That might not be true for some \epsilon (|x-1|<\delta<1. I might be missing something big here, thank you in advance.

 

[dzogi]

I think I sorted this out. We're taking \delta<1, so it doesn't matter since we only have to find one \delta to prove that the limit holds.

 

[HallsofIvy]

I'm trying to understand \epsilon-\delta proofs, but I'm having some trouble. For example, if we want to prove that \lim_{x\rightarrow2}x^3=8, starting from |x^3-8| we get to something like

|x-2||x^2+2x+4|

And this is what confuses me: we conjecture that |x-2|<1, then |x|<3, so we get

|x-2||x^2+2x+4| \leq |x-2|||x|^2+2|x|+4|<|x-2||3^2+2*3+4|=19|x-2| then we can easily find out what \delta is. But, why do we make that assumption when we're working with every \epsilon>0? That might not be true for some \epsilon (|x-1|<\delta<1. I might be missing something big here, thank you in advance.

?? It is not relevant what \epsilon is, you are requiring that \delta be less that 1. That, you are free to choose. As long as there is some \delta works, then any smaller \delta also works. That's why we can require that \delta be less than 1.

In this particular case, what we would have to do is choose \delta less than the smaller of \epsilon/19 and 1.