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

  1. #1
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    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.

    I Still Don't Understand Epsilon-Delta Proofs-capture.png


    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..)
    Last edited by lc99; Oct 1st 2017 at 09:04 PM.
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  2. #2
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    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

    $\displaystyle \lim_{x\to 2} 3x-5 = 1$
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  3. #3
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    Re: I Still Don't Understand Epsilon-Delta Proofs

    Thanks from lc99
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  4. #4
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    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.

    I Still Don't Understand Epsilon-Delta Proofs-capture.png
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  5. #5
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    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:

    $|x-0| < \varepsilon^2 \Longrightarrow \sqrt{|x-0|} < \varepsilon$
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  6. #6
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    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
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