# Epsilon-Delta proofs, once again

• dzogi

#### 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.

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.

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.