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Basic limit proof of limit equivalence

  1. Oct 22, 2013 #1
    1. The problem statement, all variables and given/known data

    Prove that limx[itex]\rightarrow[/itex]cf(x)=L if and only if limh[itex]\rightarrow[/itex]0f(x+h)=L.

    2. Relevant equations



    3. The attempt at a solution

    I think this is a simple problem, but I am getting caught up in the middle, as I'm not sure if my procedure is a valid way to prove the statement.

    Suppose limx[itex]\rightarrow[/itex]cf(x)=L.

    Then for each [itex]\epsilon[/itex]>0 there is a [itex]\delta[/itex]>0 so that

    if |x-c|<[itex]\delta[/itex] then |f(x)-L|<[itex]\epsilon[/itex].

    Let x=c+h. Then if |x-c|<δ [itex]\Rightarrow[/itex] |c+h-c|<δ [itex]\Rightarrow[/itex] |h|<δ. Furthermore, |f(c+h)-L|<ε. And since we assumed limx[itex]\rightarrow[/itex]cf(x)=L, it follows then that limh[itex]\rightarrow[/itex]0f(c+h)=L.*

    So my questions is this: it is valid to simply let x=c+h as I did?

    *Note, I did not prove the other case, but I just wrote this out so you guys can get a good idea of what my argument is and tell me if it is wrong or not.
     
  2. jcsd
  3. Oct 22, 2013 #2

    Office_Shredder

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    This is the right approach, but your arrows don't point the way you want them to. To show that if h goes to zero, f(c+h) goes to L, you need to start with: if |h| < δ, things happen. You started by assuming that |x-c| < δ, then showed for your choice of defining h that |h| < δ as well, which isn't good enough. But all the stuff you wrote down is reversible, so you should be able to start with |h| < δ and then use what you know about f(x) when x=c+h.
     
  4. Oct 22, 2013 #3
    Thank you very much Office_Shredder! One question though: I thought that I had to prove it in both orders, since it is an if and only if statement. Your route certainly seems more sensible but if I were to write the complete solution, wouldn't I assume in one case that as h goes to zero f(c+h) goes to f(c) and show that f(x) going to f(c) follows, and then start over and assume that as x goes to c f(x) goes to f(c) and show that f(c+h) going to f(c) follows from there? What I am asking is if I wanted to prove A iff B, wouldn't I do:

    If A is true, then B follows.
    If B is true, then A follows.

    Thus, A is true iff B is true. ? (Hopefully that example makes more sense than the above).

    Again, thank you!
     
  5. Oct 22, 2013 #4

    Office_Shredder

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    Yes, but it's important to prove if A then B in the section where you are claiming if A then B. If you want to assume
    [tex] \lim_{x\to c} f(x) = L [/tex]
    and then prove that
    [tex] \lim_{h\to 0} f(c+h) = L [/tex],
    then you better at some point have h be an arbitrary number such that |h| < δ
     
  6. Oct 22, 2013 #5
    Ohhhhh! I see. Thank you!
     
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