Thread: I Still Don't Understand Epsilon-Delta Proofs

1. I Still Don't Understand Epsilon-Delta Proofs

Hey guys. Sorry if i dont make sense because i'm just trying to make sense of this concept !

Ok, so I understand that to prove that a limit exists that there HAS to be a delta value (S) where the distance between x and c is less than S which satisfies that distance between f(x) and L is less than E.

So basically if i can find a delta that is greater than 0 and that if i plug in x+delta, the distance f(x) and L is less than epsilon (sorry if im repeating but im still not sure...)

Also, I understand the graphical approach. That is, if i pick any epsilon value , i should be able to get a distance between delta value (S) and x that is within that distance between f(x) and L.

Anyway, i THINK i understand the conceptual part, but im confused with the actual proof/scratch work, and how it's proving what i said above^

Example:

How does the proof below show that? How does solving for abs( f(x) - L ) = E tell us that this is true. I don't understand any of this, and I've been trying to understand this for hours and watching videos, but it just gets soo confusing.

It doesn't make sense to me at all and i feel hopeless. I need to go to bed now, but I've been trying to really understand this, and it's not going too well. (I already looked at the epsilon-delta notes from the calc forum but still confused..)

2. Re: I Still Don't Understand Epsilon-Delta Proofs

You're not the first person utterly frustrated by $\delta-\epsilon$ proofs

Let's go through the definition once more.

$\displaystyle \lim_{x\to c} f(x) = L$

means

$\forall \epsilon > 0$

for any positive $\epsilon$ you give me

$\exists \delta > 0$

there exists another positive number $\delta$

$\ni |x-c|< \delta \Rightarrow |f(x)-L| < \epsilon$

such that if $x$ is closer to $c$ than $\delta$ then $f(x)$ is closer to $L$ than $\epsilon$

now show $\displaystyle \lim_{x\to 2} (3x-5) = 1$

given $\epsilon > 0$ we need to find $\delta > 0$ such that

$|x-2|<\delta \Rightarrow |3x - 5 - 1| < \epsilon$

$|3x-5-1| < \epsilon$

$|3x-6|< \epsilon$

$|x-2| < \dfrac{\epsilon}{3}$

so let $\delta = \dfrac{\epsilon}{3}$

\begin {align*} |x-2|<\delta &\Rightarrow \\ |x-2| < \dfrac{\epsilon}{3} &\Rightarrow \\ |3x-6|< \epsilon &\Rightarrow \\ |3x-5 -1| < \epsilon \end{align*}

and combining all that

$|x-2| < \delta \Rightarrow |3x-5-1|< \epsilon$

and thus

$\displaystyle \lim_{x\to 2} 3x-5 = 1$

4. Re: I Still Don't Understand Epsilon-Delta Proofs

Thanks! It makes a bit more sense the way you wrote it, but I'm trying to prove this one example of a limit and it is confusing me.

5. Re: I Still Don't Understand Epsilon-Delta Proofs

Start with what you want to prove, try to find a viable delta.

If we want $\sqrt{|x|} < \varepsilon$, then $|x| < \varepsilon^2$. So, $|x-0| < \varepsilon^2$. We now have a delta that will work.

Given any $\varepsilon>0$, let $\delta = \varepsilon^2$. For any $x \in \mathbb{R}$ such that $|x-0| < \delta$, we have:

$|x-0| < \varepsilon^2 \Longrightarrow \sqrt{|x-0|} < \varepsilon$

6. Re: I Still Don't Understand Epsilon-Delta Proofs

Thanks for breaking it down. I think i actually understand these now. Thanks so much again