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Prove a limit using epsilon-delta definition

  1. Sep 8, 2016 #1

    Math_QED

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

    Proof that:

    ##\lim_{x \to 1} \frac{1}{1+x} = \frac{1}{2}##

    using the epsilon-delta definition of a limit.

    (Problem below)

    2. Relevant equations

    ##\lim_{x \to 1} \frac{1}{1+x} = \frac{1}{2} \iff \forall ε>0, \existsδ>0, \forall x ∈ \mathbb{R}\backslash\{-1\}: 0 < |x-1| < δ \Rightarrow |\frac{1}{1+x} - \frac{1}{2}|<ε##

    3. The attempt at a solution

    Draft

    So, I have to find δ(ε) such that:

    ##|x-1| < δ \Rightarrow |\frac{1}{1+x} - \frac{1}{2}|<ε##
    and, after simplifying I got:
    ##|\frac{1}{1+x} - \frac{1}{2}|<ε \Rightarrow |x-1| < ε|2x+2|##

    Now, I chose ##δ = 1##, then ##|x-1| < 1##, thus ##2< 2x + 2 < 6##. So I decided to take ##δ = min\{1,6ε\}##

    Proof

    We take ε arbitrary and we choose ##δ = min\{1,6ε\}##

    Case 1: δ = 1, then 6ε > 1 as ##δ = min\{1,6ε\}##.
    We then have, ##|x-1|<δ \Rightarrow |x-1| < 1 \Rightarrow |x-1| < 6ε \Rightarrow \frac{|x-1|}{6} < ε##

    Now, here comes the problem. I wanted to use that ##6>|2x + 2|## (this result was obtained in the draft). However, when I would do this, the total fraction will become larger (as the denominator becomes smaller), and it would be possible that the inequality does not hold anymore. I got a similar problem with case 2. I'm learning this on my own, so it is possible that I'm totally wrong. Thank you for helping me in advance. Also, posting in Latex for the first time. Any hints on this matter would also be appreciated.
     
    Last edited: Sep 8, 2016
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  3. Sep 8, 2016 #2

    mfb

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    The 6 ε are too large, you'll need something smaller.
    There is no need to include prefactors larger than one in general. It does not improve anything, it just makes the inequalities harder to satisfy.
     
  4. Sep 8, 2016 #3

    Math_QED

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    How can I find what delta is better to pick?
     
  5. Sep 8, 2016 #4

    PeroK

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    You started off well, but then you took a wrong turn. Try starting from:

    ##|\frac{1}{1+x} - \frac{1}{2}| = |\frac{1-x}{2(1+x)}|##

    Also, in these proofs, I would leave out the epsilons and deltas until I'm ready. Bringing them in at the start is not always a good idea.
     
  6. Sep 9, 2016 #5

    epenguin

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    Yes I think you just calculated |x - 1| when correct calculation would have given you |1 - x| .

    But I want to ask the professional mathematicians and teachers here, it seems to me this shows something.

    Firstly this concept of limits is supposed to be a difficult one OK? And that is the reason in order to get it across, students are made to do tons and tons of exercises on it. But actually it looks to me that really their difficulty is usually not concepts of limits, but the ordinary algebraic manipulations in doing these calculations! As in this case. Whether then they 'get it' in a way that stays as a result of that drilling I don't know, do you?

    And look at this – above you have an expression which is obviously ½ when x = 1 There is nothing peculiar about it. It is not looking like 0/0 or anything weird like that. Can you make students who are going to be at most ornery users of math like engineers plough through this? Which seems a bit too much of a hoo-ha to me.

    Ah but, I expect you will say this is just to get them used to easy examples and then we will present them with some things that are weird and do look like 0/0. Except that then, as I said, you get the students going crazy on the supposedly easy examples! Maybe things like
    (sin x)/x are a bit more fun. But that I can think of, even those who go on to be regular users of math are rarely going to meet these things. In fact that I can recall there is only one place a lot of people meet it usually, that is in the basic thing of calculus where you say (lim δx →0 of) [f(x + δx) - f(x)]/δx and top and bottom do look to be 0. But then for generations I think that one was explained in just minutes, part of one lesson, and people took it in their stride.

    Maybe you could say that limits come a bit more into their own with sums of infinite series - okay it may not be strictly quite right to say they ever add up to something and you should say is their sum has that number as limit. But I mean to say even this is such a big deal?

    When I was at school, mnya mnyah, I don't remember we had any of this except the above examples. At my one math year in university we did, and it wasn't any problem for me. And just because of the relief of a thing which wasn't a problem I never questioned it. But if people do have a big problem then perhaps we should?

    Just askin'. :olduhh:
     
  7. Sep 9, 2016 #6

    Mark44

    Staff: Mentor

    |x - 1| and |1 - x| are equal for all values of x.
    If you're proving the limit of a function whose graph is a straight line, the proof is pretty easy. As soon as there's a curve involved, things get more difficult.
    The business with ##\delta## and ##\epsilon## brings rigor, rather than hand-waving, to the argument.
     
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