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Please help with seemingly easy limit proof

  1. Jul 6, 2011 #1
    1. The problem statement, all variables and given/known data

    Let f: R ---> R, c in R.
    Show that lim f(x) (as x goes to c) = L iff lim f(x+c) (as x goes to 0) = L



    2. Relevant equations

    n/a



    3. The attempt at a solution

    First of all, I understand why this is true conceptually, but I'm having trouble writing a rigorous proof. I think if somebody could just help me get started on the forward direction, then i can take it from there.

    There's two methods I could use: (1) I could use the sequential criterion for limits or (2) I could use the definition of the limit.

    For (1), Suppose lim f(x) (as x goes to c) = L. Then by the sequential criterion, for every sequence (x(n)) contained in R such that (x(n)) converges to c, but x(n) /= c for all n, then the sequence (f(x(n))) goes converges to L. That's fine, but I'm having trouble explaining how this implies that for sequences (x(n)) that converge to 0, then (f(x(n)+c)) converges to L. Do i need to define a separate sequence converging to 0, then use the Algebraic Theorem for sequences? I don't know how to put this rigorously.

    For (2), Suppose the same hypothesis as (1). Then, for all e>0, there exists a d > 0, such that if 0 < |x - c| < d, then |f(x) - L| < e. I can use the reverse triangle inequality to show that |x| < d + |c|, then define a new delta, call it d2 = d + |c|. Then i will have the condition that |x - 0| < d2. But I'm having trouble showing how this implies |f(x+c) - L| is less than some new eps, call it e2. How do I combine f(x) and c, into f(x+c)? Is there some theory of linearity that i can use for adding constants to functions defined on R?


    Any help with the rigorous argument would be really awesome. Which method should i use , the sequential criterion or the definition of limit? Like I said, i understand the concept completely ( i could draw a pictorial proof easily), I'm just having trouble putting it on paper into a rigorous proof that will convince my analysis instructor that I know what I'm doing :)
     
  2. jcsd
  3. Jul 6, 2011 #2
    Try letting [itex] u = x+c [/itex] and then noting [itex] |x| = |(x+c) - c| = |u-c|[/itex].
     
  4. Jul 6, 2011 #3
    Thanks, that was really helpful! So how's this look for the forward direction?


    Suppose lim f(x) (as x goes to c) = L. Let e > 0. Then, by hypothesis, there exists d > 0 such that 0< |x - c| < d implies |f(x) - L| < e. Now, let u := x - c. Thus 0 < |u| < d implies |f(u + c) - L| < e. Hence, by the definition of limit, we have lim f(u +c) (as u goes to 0) = L.

    Now i want to conclude that since u is arbitrary, then lim f(x+c) (as x goes to 0) = L. But can I really do that? Is u really arbitrary? It was defined very specifically, after all. I'm still having trouble understanding this subtlety. x and u are different, but they just represent arbitrary variables, so ultimately, they fundamentally represent the same thing? Is this right?
     
    Last edited: Jul 6, 2011
  5. Jul 6, 2011 #4
    why don't you plug in c = x - u? Wouldn't that work?
     
  6. Jul 6, 2011 #5
    I'm not following you, that's basically the same thing that i did, isn't it?

    I'm pretty sure this proof is good (yes? no? let me know!), I'm just trying to wrap my head around the fact that saying lim f(u+c) (as u goes to 0) = L is equivalent to saying lim f(x+c) (as x goes to 0) = L, in light of the way that u was defined.
     
  7. Jul 6, 2011 #6
    Here, I'll do one implication in detail.

    Assume for all [itex] u \in \mathbb{R}[/itex] for all [itex] \epsilon >0 [/itex] there is a [itex]\delta >0[/itex] such that if [itex]0 < |u -c | < \delta [/itex] then [itex]|f(u) - L| < \epsilon[/itex].

    Fix [itex] \epsilon >0 [/itex] and take [itex] \delta [/itex] as in the above assumption. Suppose [itex]x \in \mathbb{R}[/itex] with [itex] 0<|x| < \delta[/itex]. Let [itex] u = x+c[/itex]. Then [itex]|x| = |(x+c)-c| = |u-c|[/itex] so [itex]0 < |u -c | < \delta[/itex]. By assumption, then [itex] |f(u) -L| <\epsilon[/itex], so [itex] |f(x+c) -L| <\epsilon[/itex]. Since x was arbitrary, then for any x with [itex] 0<|x| < \delta[/itex], [itex] |f(x+c) -L| <\epsilon[/itex]. Therefore [itex] \lim_{x \to 0} f(x+c) = L [/itex].

    The only thing that is arbitrary in the above proof is the x I picked in the first line (well, and epsilon, too).

    Can you do the other direction or justify your proof now?

    Maybe I'm misunderstanding you, but we don't get to choose c, it's given to us.
     
    Last edited: Jul 6, 2011
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