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Two more limit problems

  1. Jul 8, 2012 #1
    Problem 1
    "Prove that lim[x->2] 1/x = 1/2." You have to do it using the epsilon-delta-definition of a limit. And the truth is, I don't 'get' those epsilon-delta limits at all (except the easiest kind. Sort of). My book (Stewart's Calculus) either doesn't explain them very well, or I'm just missing something obvious. Can anyone help me here?

    Problem 2
    "If the function f is defined by f(x) = 0 if x is rational; 1 if x is irrational, prove that lim[x->0] f(x) does not exist." Ehm, here I have simply no idea where even to start. My guess would be that there are an infinite number of rational and irrational numbers close to zero, which means f(x) infinitely oscillates, but how would one go about proving this? Is my guess even right?
     
  2. jcsd
  3. Jul 8, 2012 #2

    tiny-tim

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    Hi KiwiKid! :smile:
    your guess is right :smile:

    you have to prove that, no matter how small the neighbourhood of 0, it contains two different xs with f(x) = 0 and f(x) = 1 :wink:
    you have to prove that, no matter how small the neighbourhood of 1/2, you can find a neighbourhood of 2 such that f of that whole neighbourhood lies in the first neighbourhood :wink:
     
  4. Jul 8, 2012 #3

    Zondrina

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    I'll help you with your first one.

    You want to use this definition for this :
    [itex]\forall[/itex]ε>0, [itex]\exists[/itex]δ>0 | 0 < |x-a| < δ [itex]\Rightarrow[/itex] |f(x)-L| < ε

    So let us sub in what we know :
    [itex]\forall[/itex]ε>0, [itex]\exists[/itex]δ>0 | 0 < |x-2| < δ [itex]\Rightarrow[/itex] |1/x-1/2| < ε

    So we want to find a δ which will satisfy this definition. I've done this problem before and im pretty sure δ=min{1,2ε} is what we are going to get. Lets find out!

    So we start by working with our |f(x)-L| < ε :
    |1/x - 1/2| = |(2-x)/2x| = (1/2)*|2-x|*1/|x|

    Now remember that :
    |2-x| = |(-1)(x-2)| = |-1||x-2| = 1|x-2| = |x-2|

    We also know that |x-2| < δ so using these two facts we continue forth :
    |1/x - 1/2| = |(2-x)/2x| = (1/2)*|2-x|*1/|x| < (1/2)*δ*1/|x|

    Now we want to get rid of that pesky |x| so we can find our δ. So we ask ourselves, what do we know about |x|? Well we know |x-2| < δ and we can manipulate this to give us what we desire. It also helps at this point to give a bound on δ. Lets say that δ[itex]\leq[/itex]1 and observe :

    |x-2| < 1
    -1 < x-2 < 1
    1< x < 3

    So now we know that we can choose x between these bounds, but we want 1/|x| to get as large as it possibly can, so the smaller the denominator becomes, the larger the entire expression becomes. So let us pick x = 1 and observe :

    |1/x - 1/2| = |(2-x)/2x| = (1/2)*|2-x|*1/|x| < (1/2)*δ*1/|x| [itex]\leq[/itex] δ/2 [itex]\leq[/itex] ε

    This yeilds : δ [itex]\leq[/itex] 2ε and therefore δ=min{1,2ε}. Now we will prove this value of δ works no matter what.

    [itex]\forall[/itex]ε>0, [itex]\exists[/itex]δ=min{1,2ε} > 0 | 0 < |x-2| < δ [itex]\Rightarrow[/itex] |1/x-1/2| < ε

    Same steps as before :

    |1/x - 1/2| = |(2-x)/2x| = (1/2)*|2-x|*1/|x| < (1/2)*δ*1/|x| [itex]\leq[/itex] (1/2)(2ε) = ε

    So right there we have shown |f(x)-L| < ε when δ=min{1,2ε} concluding the proof.

    Hope that helps!
     
    Last edited by a moderator: Jul 9, 2012
  5. Jul 8, 2012 #4
    Well, I guess it's one thing that at least I guessed that right, but now I still don't really have a clue what to do. I know what a limit is (so unfortunately your description doesn't help me), but I don't really know how to do all this with all those deltas and epsilons. As for the first problem, I still have no clue how to prove that there are always two 'types' of numbers within 0 and delta.

    Zondrina, I appreciate your help, but, uhm, I have no idea what half of those symbols stand for.
     
  6. Jul 8, 2012 #5

    Zondrina

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    Which ones do you not understand?
     
  7. Jul 8, 2012 #6
    The upside-down-A, the mirrored E, and the '|'. Also, I'm not completely sure I understand the 'min' function. Are you allowed to just put that in a proof?

    I'm sorry, I'm not yet very comfortable with these mathematical proofs. I'm used to problems where you get one clear answer, not answers that depend on three different symbols and functions at the end. :/
     
  8. Jul 8, 2012 #7

    tiny-tim

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    well, that's easy … you have to prove that there's at least one rational number and one irrational number in (-δ,δ) :wink:
    you choose a δ,

    then you have to define an ε which depends on δ

    (usually it's something like ε = δ/2 or ε = δn) :smile:
     
  9. Jul 8, 2012 #8
    Well, I *know* that those numbers exist, but like I said, I have very little experience with mathematical proofs. 'Prove' that there's at least one of each number? How does one do that mathematically?

    Wait, what? So far I've heard that you choose an epsilon, and then define a delta that depends on it. Or is it the same thing? And how can you even know which one to 'choose'?
     
  10. Jul 8, 2012 #9

    tiny-tim

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    i'd be inclined to say that it's so obvious that you can just state it without proof

    but if you need a proof, i'd start by saying that 1/δ will be large, but there'll always be a whole number N greater than 1/δ, and then 1/N is a rational number less than δ

    then you need an irrational number … i'd go for a square root of a whole number :wink:
    i can never remember which way round it is :rolleyes:: if i've got it the wrong way, then just swap them round :smile:

    (and to find what to choose, study some of the example proofs in your book)
     
  11. Jul 8, 2012 #10

    Zondrina

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    The upside down A means "For all"
    The Backwards E means "There exists"
    And the | means "Such that"

    For example :
    ∀ε>0, ∃δ>0 | 0 < |x-a| < δ ⇒ |f(x)-L| < ε

    Is read :

    For all epsilon greater than zero, there exists delta greater than zero such that 0 less than the absolute value of x-a less than delta implies the absolute value of the function minus its limit is less than epsilon.

    Also min means that you are choosing the minimum value from that set which is required. In these kinds of proofs you will see it a LOT.
     
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