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Thread: Using epsilon-delta definition of a limit

  1. #1
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    Using epsilon-delta definition of a limit

    Hi there
    I'm doing a worked example in a Larson book. I don't understand some of it. I have scanned this excerpt from Larson.
    Using epsilon-delta definition of a limit-limitscropped.jpg


    So what I don't understand is "Thus, letting \delta be the minimum of \varepsilon/5 and 1, it follows that ....."
    I don't understand why he is saying letting delta be the minimum of epsilon/5 and 1, and where he is getting 1 from? I'm thinking 1 is related to the solution of |x - 2| in the interval (1,3). In that interval the largest number that |x-2| can evaluate to is 1, or almost 1 and when that happens |x+2| = 5 or almost 5 because it's an open interval.
    Am I thinking correctly? And does anybody else have anything to add to enhance my understanding becauuse even if I'm thinking right there, I still don't feel fully understand what he is doing in this example.
    Dave.
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  2. #2
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    Re: Using epsilon-delta definition of a limit

    Do you under why any of this is done. We want to make the distance between $4$ and $x^2$, $|x^2-4|$, as small as necessary when $x$ is close to $2$, that is the distance $|x-2|$ small enough to make that happen. Because $x\to 2$ you can control the distance $x$ is from $2$.

    Now Larson used 1, let use use 3. So we start with $|x-2|<3$, $x$ is within three units of $2$.
    Basic algebra: $\left\{ \begin{array}{l}|x-2|<3\\-3<x-2<3\\-1<x<5\\-7<1<x+2<7\\|x+2|<7\end{array} \right.$
    The last step may take some thought.

    Most students can follow algebra, even the fact $-7<1$ is true, but WHY go after $|x+2|~?$
    The answer is so mundane: $|x^2-4|=|x+2|\cdot |x-2|$ that is why. We want to make that as small as possible.

    If we insure that $|x-2|<3$ we know that $|x+2|<7$.
    Thus take any positive $c>0$ if we say $d=\min\{c/7,3\}$ then $|x-2|<d\le 3$ insures that $|x+2|<7$.

    Put that together to get $|x^2-4|=(|x-2|)(|x+2|)<(c/9)(9)<c$ DONE.
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  3. #3
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    Re: Using epsilon-delta definition of a limit

    I am still trying to digest this to be honest. I tink I need to go and do some more reading and come back to this, and I am totally confused as to where the 9s came from.
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  4. #4
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    Re: Using epsilon-delta definition of a limit

    I can understand everything up to "Thus take an positive c > 0......"
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    Re: Using epsilon-delta definition of a limit

    Quote Originally Posted by Bucephalus View Post
    I am still trying to digest this to be honest. I tink I need to go and do some more reading and come back to this, and I am totally confused as to where the 9s came from.
    It could be any number greater or equal to then seven. Once again, we need to bound the term $|x+2|$ because we have control over the term $|x-2|$.

    Quote Originally Posted by Bucephalus View Post
    I can understand everything up to "Thus take an positive c > 0......"
    It is part of the definition: $\displaystyle{\lim _{x \to a}}f(x) = L$ means
    if $c>0$ there is a positive number $d$ such that if $|x-a|<d$ then $|f(x)-L|<c.$
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    Re: Using epsilon-delta definition of a limit

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    Re: Using epsilon-delta definition of a limit

    Quote Originally Posted by Plato View Post

    It is part of the definition: \displaystyle{\lim _{x \to a}}f(x) = L means
    if c>0 there is a positive number d such that if |x-a|<d then |f(x)-L|<c.
    Yes i'm familiar with this definition. I'm just not getting the d=min{...} stuff, but I think it's only because I'm not conceptualising the |x-2| < d relationship with c yet. I have to go and concentrate, but before I do that I will go and read that document that Jeff posted.
    Last edited by Bucephalus; Apr 11th 2017 at 07:27 AM.
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  8. #8
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    Re: Using epsilon-delta definition of a limit

    Quote Originally Posted by JeffM View Post
    Thanks JeffM for the reference. It filled in all the holes that were troubling me.
    Thanks Plato for helping me out too.
    Dave.
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