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Help with proof of the uniqueness of limits.

  1. Sep 20, 2004 #1
    Good evening,
    I am a first year engineer here and a first time poster also.

    I had a problem that has been bugging me for the last few days; after much head-scratching and tree-killing, I may have solved it. I am, however, not sure at all if all my assumptions along the way are correct. So I am here to seek wisdom.

    The question deals with proving the uniqueness of limits.

    Prove that all limits are unique.

    The textbook got me started, it said to define
    [tex]\lim_{x\to a} f(x) = L [/tex]
    [tex]\lim_{x\to a} f(x) = M[/tex]

    Assume [itex]L \neq M [/itex] and let [itex]\frac{|L - M|}{3} = \epsilon[/itex]

    So here goes my proof..

    [tex]
    \begin{align*}
    &|L-M| = |L -M + f(x) - f(x)| = |(L-f(x)) + (f(x) - M)| \\
    &|L-M| \leq |-1||f(x) - L| + |f(x) - M| = |f(x) - L| + |f(x) - M|
    \end{align*}
    [/tex]

    So by definition..
    [tex]
    \begin{align*}
    0 \leq |x-a| \leq \delta (\epsilon) \implies \substack{|f(x) - L| \leq \epsilon \\ and \\ |f(x) - M| \leq \epsilon} \\
    \end{align*}
    [/tex]
    [tex]
    \begin{align*}
    \therefore |L-M| &\leq 2\epsilon \\
    |L-M| + \epsilon &\leq 3\epsilon \\
    |L-M| + \epsilon &\leq 3 \left(\frac{|L-M|}{3}\right) \\
    |L-M| + \epsilon &\leq |L-M|
    \end{align*}
    [/tex]

    Which can not possibly be true, so I conclude that our initial assumption [itex]L \neq M[/itex] was false, and therefore L must equal M.

    Is this not the way to answering the question? If not, how should I look at this problem? What should I have done differently? What other approaches should I take?

    (Side note... I REALLY hate delta-epsilon proofs..)
     
    Last edited: Sep 20, 2004
  2. jcsd
  3. Sep 20, 2004 #2

    arildno

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    "So I am here to seek wisdom."
    Seek and ye shall recieve! (Welcome to PF!)
    Yep, you've got it.
    Nope, those proofs are cuties..
     
  4. Sep 20, 2004 #3
    So this is "hand-in"able? :P

    Thanks for the timely reply, by the way.
     
  5. Sep 20, 2004 #4

    mathwonk

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    It is much easier in words. i.e. if L is the limit of f(x) as x approaches a, then the inverse image of every interval centered at L contains a punctured interval centered at a. same for M. But this is a contradiction if Ldoes not equal M, since then L,M are centers of disjoint intervals whose inverse images are thus disjoint. But no two punctured intervals centered at a are disjoint.

    the only place epsilon is needed is to describe the size of the disjoint intervals centered at L,M, namely |L-M|/3.
     
    Last edited: Sep 21, 2004
  6. Sep 21, 2004 #5
    Alright, this also makes sense, doesn't it?

    We ignore the assumption [tex]L \neq M[/itex] and [tex]\epsilon = \frac{|L-M|}{3}[/itex]
    [tex]
    \begin{align*}
    &|L-M| = |L -M + f(x) - f(x)| = |(L-f(x)) + (f(x) - M)| \\
    &|L-M| \leq |-1||f(x) - L| + |f(x) - M| = |f(x) - L| + |f(x) - M| \\
    &|L-M| \leq |f(x) - L| + |f(x) - M|
    \end{align*}
    [/tex]
    [tex]
    \therefore |L-M| \leq 2\epsilon
    [/tex]

    And since [itex]\epsilon[/itex] is can be as arbitrarily small as we want it to be, [itex]L-M[/itex] must equal 0. Is this also correct?

    Side note: When using [itex]\delta - \epsilon[/itex] proofs, do we use [itex] \leq [/itex] or [itex] < [/itex], or does it not matter? My prof uses the former, and the textbook uses the latter.

    Can I still assume [itex]|L-M| = 0[/itex] if I use [itex]\leq[/itex] ? Or do I have to have it strictly less than [itex]\epsilon[/itex]?
     
  7. Sep 21, 2004 #6

    mathwonk

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    that looks nice. as to your question, ask your self: is it true that if L is such that
    0<= L <= a, for all positive a, then L is zero?

    also: is it true that if L is such that
    0<= L > a, for all positive a, then L is zero?
     
  8. Sep 21, 2004 #7
    If by the second statement you meant [itex]0 \leq L < a[/itex] then yeah, I guess they are equivalent statements.

    Thank you for your help.
     
  9. Sep 21, 2004 #8

    mathwonk

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    right you are
     
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