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Epsilon delta proofs equaling a constant

  1. Sep 14, 2012 #1
    1. The problem statement, all variables and given/known data

    Lim x→a of f(x) = c (Where c is a constant)

    2. Relevant equations



    3. The attempt at a solution
    I have no idea. I am able to do these if I can manipulate fx-L to equal x-a but I am having trouble with this one. Please help me!
     
  2. jcsd
  3. Sep 14, 2012 #2

    LCKurtz

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    I have no idea either. What is your question? What are you trying to do?
     
  4. Sep 14, 2012 #3
    I have to prove using the epsilon delta method that as x approaches a the limit of a constant is a constant. It seems so easy to do that I'm having trouble with it.
     
  5. Sep 14, 2012 #4

    SammyS

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    δ can be pretty much anything.

    Let δ = 1, for example.
     
  6. Sep 14, 2012 #5
    I understand that but my proff wants us to mathematically prove why that is. Wich seems difficult because it is s simple.
     
  7. Sep 14, 2012 #6

    HallsofIvy

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    You are claiming that for f(x)= c, a contant, for any a, [itex]\lim_{x\to a} f(x)= c[/itex]. The definition of "[itex]\lim_{x\to a} f(x)= L[/itex]" is "Given [itex]\epsilon> 0[/itex], there exist [itex]\delta> 0[/itex] such that if [itex]|x- a|< \delta[/itex] then [itex]|f(x)- L|< \epsilon[/itex]". Here, of course, f(x)= c and L= c so that [itex]|f(x)- L|= |c- c|= 0<\epsilon[/itex] for all x. So say
    "Given [itex]\epsilon> 0[/itex], let [itex]\delta[/itex] be any positive number. Then if [itex]|x- a|< \delta[/itex]... (you do the rest)."
     
    Last edited by a moderator: Sep 14, 2012
  8. Sep 14, 2012 #7
    Thank you very much :) after which I say since [itex]\left|x-a \right|[/itex] = [itex]\left| fx-L \right|[/itex] then [itex]\left| fx-L \right|[/itex] > δ. And that completes the proof?
     
  9. Sep 14, 2012 #8
    No, what I wrote can't be right.. Uh
     
  10. Sep 14, 2012 #9

    SammyS

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    How about:

    For ε > 0

    If | x - a | < δ

    then | f(x) - c | =   ?  
     
    Last edited: Sep 14, 2012
  11. Sep 14, 2012 #10
    All the comments have been very helpful and I'm very embarrassed because I'm very good at math.. I think I'm really overthinking it. Can someone just post the solution so I can follow the steps and really learn this?
     
  12. Sep 14, 2012 #11

    Mark44

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    Yes. Don't overthink it.
    Not going to happen.

    See the rules (https://www.physicsforums.com/showthread.php?t=414380), especially this one:
     
  13. Sep 14, 2012 #12
    Guess I should have read that first. It has to be l fx-c l < δ that's what I'm thinking
     
  14. Sep 14, 2012 #13

    Mark44

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    No, you want to show that |f(x) - C| < ##\epsilon##. That really shouldn't be hard to do, given the function you're working with.
     
  15. Sep 14, 2012 #14
    SInce l fx-L l = 0 and ε > 0 then l fx - L l < ε. Please be right....
     
  16. Sep 14, 2012 #15

    LCKurtz

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    And notice that, for this particular function, it doesn't matter how big or small ##\delta## is.
     
  17. Sep 14, 2012 #16
    Yeah it finally clicked. If l fx - L l is 0 and δ is any positive number, Given the information, it fx-L must be less than
     
  18. Sep 14, 2012 #17

    Mark44

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    You really should get into the habit of writing function notation correctly. It's f(x), not fx.
     
  19. Sep 14, 2012 #18
    Here an example
    http://imageshack.us/a/img708/1055/10204791.jpg [Broken]
     
    Last edited by a moderator: May 6, 2017
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