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Trying to understand epsilon-delta defn of limit

  1. Oct 17, 2009 #1
    I made this question up myself, so I'm not even entirely sure if it makes sense, but I'm pretty sure that it does.

    1. The problem statement, all variables and given/known data

    Consider the function
    [tex]
    \begin{equation*}
    f(x) = \left\{
    \begin{array}{cc}
    1 & : x = \dfrac{1}{n}, n \in \mathbb{N}\\
    0 & : x \not = \dfrac{1}{n}, n \in \mathbb{N}
    \end{array}
    \end{equation*}
    [/tex]

    Prove that
    [tex]
    \begin{equation*}
    \lim_{x \to \frac{1}{n}} f(x) = 0
    \end{equation*}
    [/tex]

    2. Relevant equations



    3. The attempt at a solution

    I'm really new to all this epsilon delta stuff, so I'm not sure how I would get started on a question like this.

    Given [tex]\epsilon>0[/tex] we want to show that there exists [tex]\delta > 0[/tex] such that

    [tex]
    0<|x-\dfrac{1}{n}|<\delta\\
    \implies |f(x)-0| < \epsilon
    [/tex]

    Like I said, I made this question up myself so I could try to understand this. I don't know how to get started though. Any hints?
     
  2. jcsd
  3. Oct 17, 2009 #2
    As x-->(1/n), f(x) = 0, which means your inequality holds for any epsilon. And regardless of if the value of delta(even 0) the limit exist.

    When we talk about limits we are talking about the neighborhood around your point. Your function is zero everywhere except for a bunch of rational numbers between 0 and 1.

    I don't like the question because I believe it accomplishes the exact opposite of what you want.

    Try using a simpler example then work your way up.

    The attached pdf below might be useful to you.
     

    Attached Files:

  4. Oct 17, 2009 #3
    well, it is a weird example, but i'd still like to understand it. i have a good intuition for how it works when it comes to continuous functions like polynomials... it's the weird examples that i'm trying to wrap my head around.

    i just want to figure out what a rigorous proof of this would look like.
     
  5. Oct 17, 2009 #4
    If you wanted to write a formal proof, you could note that [tex]\frac{1}{n - 1} > \frac{1}{n} > \frac{1}{n + 1} [/tex]. This gives you explicit bounds on where delta can be for a given n.
     
  6. Oct 17, 2009 #5

    Hurkyl

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    Gold Member

    More detail is needed. I assume:
    . f is supposed to be a function of a real-number variable
    . n denotes a positive integer
    . You want to prove this for all n
    . the limit is to be taken as x varies over real numbers near 1/n

    In that case, the limit is, in fact, 0, just as you speculate.


    Why is it zero? The thing to "understand" is that your function f is zero everywhere near 1/n -- and you presumably understood that, which is why you made the question up!


    So, to make a rigorous proof, the thing you need to do is to translate your understanding into things you can work with mathematically. The key is to explain what you mean by "f is zero everywhere near 1/n", which leads into aPhilosopher's comment.


    Incidentally, if a is any real number other than zero, then
    [tex]\lim_{x \rightarrow a} f(x) = 0[/tex]​
    Can you tell what happens in the case that a=0?



    Incidentally, if your understanding of the problem is different than what I suggested above, then say so -- and we can try to turn that into a proof.
     
  7. Oct 17, 2009 #6
    hmm...thanks for your replies everyone.

    i guess i need to be more clear with my function, but yes, that's what i had in mind.

    this is what i think so far:

    note that for all x not equal to 1/n in

    1/(n+1) < x < 1/(n-1)

    that f(x) = 0.

    The distance between 1/n and 1/(n+1) is

    1 / (n)(n+1)

    and the distance between 1/n and 1/(n-1) is

    1/ (n)(n-1)

    Suppose we take delta to be the minimum of these two distances, so that delta is equal to 1 / (n)(n+1). this implies that for x such that 0 < x < delta, |f(x) - 0| = 0 < epsilon for all positive epsilon.

    (sorry i got lazy with the tex. but i think i'm onto something here.)
     
  8. Oct 17, 2009 #7
    How would I prove that the limit as x -> 0 does not exist?

    I can see that no matter how small we choose delta, f(x) will still either be 1 or 0, so it can't possibly have a limit. But I'm not sure how that would look in a formal proof either.
     
  9. Oct 17, 2009 #8

    Hurkyl

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    Staff Emeritus
    Science Advisor
    Gold Member

    Knowing how to negate a sentence is a useful skill.

    The negation of
    For every e>0 there exists a d>0 such that for every x 0<|x|<d implies |f(x)-L| < e​
    is
    Not for every e>0 there exists a d>0 such that for every x 0<|x|<d implies |f(x)-L| < e​
    which simplifies to
    There exists an e>0 such that not there exists a d>0 such that for every x 0<|x|<d implies |f(x)-L| < e​
    There exists an e>0 such that for every d>0, not for every x 0<|x|<d implies |f(x)-L| < e​
    There exists an e>0 such that for every d>0, there exists an x such that not (0<|x|<d implies |f(x)-L| < e)​
    There exists an e>0 such that for every d>0, there exists an x such that 0<|x|<d and not |f(x)-L| < e​
    There exists an e>0 such that for every d>0, there exists an x such that 0<|x|<d and |f(x)-L| >= e​

    (This is somewhat clearer if you write it out with logical symbols rather than in English)

    For every L, there exists an e>0 such that for every d>0, there exists an x such that 0<|x|<d and |f(x)-L| >= e​

    I suggest splitting it into two cases: L=0, and L is not zero.


    Incidentally, there is a generalization of a limit: the notion of a "limit point". It turns out as x->0, f(x) has two limit points: 0 and 1. (And because there is more than 1 limit point, the limit does not exist)

    You want to use your observation -- that near 0, f(x) attains both 0 and 1 as values.
     
  10. Oct 17, 2009 #9
    Thanks for everyone's help.
     
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