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Homework Help: Proof of Limit Laws

  1. Jan 1, 2012 #1
    Dear All,

    I need help on proving: eq0073M.gif


    According to Pauls Online Notes,
    we let ε > 0. Since [PLAIN]http://tutorial.math.lamar.edu/Classes/CalcI/LimitProofs_files/eq0075M.gif, [Broken] there's a http://tutorial.math.lamar.edu/Classes/CalcI/LimitProofs_files/eq0076M.gif such that
    eq0077M.gif



    I understand that l g(x) - L l < ε, but how do we derive l g(x) - L l < l L l / 2 ?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jan 1, 2012 #2

    SammyS

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    What does it mean (in terms of δ and ε) for [itex]\displaystyle\lim_{x\,\to\,a}\ g(x)=L\ ?[/itex]
    ...

    In particular, if you let ε = |L|/2, then you know that there is some number, call it δ1, such that whenever 0 < | x - a | < δ1,
    then | g(x) - L | < |L|/2 .
     
    Last edited by a moderator: May 5, 2017
  4. Jan 1, 2012 #3
    Hi SammyS, thanks for reply. But how do you get | L |/2 ?
     
  5. Jan 1, 2012 #4

    SammyS

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    Often, the way you come up with such quantities is to "play around" with the δ and the ε expressions, (often working backwards, so to speak, that is -- from the ε to the δ) to find a relationship between δ and ε.

    So, off hand, I don't know why he chose |L|/2 , but I haven't gone through his whole argument.

    It's just that he's using the |L|/2 for ε, so we know δ1 exists from the definition of the limit and knowing that [itex]\displaystyle\lim_{x\,\to\,a}\,g(x)=L\,.[/itex]
     
  6. Jan 1, 2012 #5

    Hi, please refer to:
    http://tutorial.math.lamar.edu/Classes/CalcI/LimitProofs.aspx( [Broken] Proof of 4)
     
    Last edited by a moderator: May 5, 2017
  7. Jan 1, 2012 #6

    SammyS

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    Yes, after looking at this (I had already found it.), maybe I should ask you "What is the role of |L|/2 in the proof?"

    BTW, there is nothing magic about |L|/2 itself. All that's needed is something less than |L|.

    He uses |L|/2 to get δ1. He uses δ1 to get a bound on 1/|g(x)| , because when you consider what needs to be proved, you realize that you must have
    [itex]\displaystyle\left|\frac{1}{g(x)}-\frac{1}{L}\right|<\varepsilon[/itex]​
    A little algebra shows that
    [itex]\displaystyle\left|\frac{1}{g(x)}-\frac{1}{L}\right|=\frac{1}{|L|}\frac{1}{|g(x)|} \left|g(x)-L\right|[/itex]​
    Looking at details in the proof, we see that δ1 is used so that there is an upper bound on [itex]\displaystyle\frac{1}{|g(x)|}[/itex]. δ2 is used so that there is a bound on |g(x)-L| .
     
    Last edited by a moderator: May 5, 2017
  8. Jan 2, 2012 #7
    May I know why must l g(x) - L l < l L l ?
     
    Last edited: Jan 2, 2012
  9. Jan 2, 2012 #8

    SammyS

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    Actually, if you look at the Notes, [itex]\displaystyle \left|g(x)-L\right|<\frac{L^2}{2}\varepsilon\,.[/itex]

    This was chosen so that
    [itex]\displaystyle \frac{1}{|L|}\frac{1}{|g(x)|} \left|g(x)-L\right|<\frac{1}{|L|}\frac{2}{|L|}\frac{L^2}{2} \varepsilon\,,[/itex]​
    whenever
    [itex]0<|x-a|<\min(\delta_1,\delta_2)\,.[/itex]​
     
  10. Jan 2, 2012 #9

    And how do we know that: l g(x) - L l < (L x L)ε/2 ?
     
  11. Jan 2, 2012 #10

    HallsofIvy

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    You seem to be completely missing the point! If [itex]\lim_{x\to a} g(x)= L[/itex], then, for any number [itex]\epsilon> 0[/itex], there exist [itex]\delta[/itex] such that if [itex]|x- a|< \delta[/itex] then [itex]|g(x)- L|< \epsilon[/itex]. Because [itex]\epsilon[/itex] can be any, we can, if we wish, take it to be L, or L/2, or [itex](L^2/2)\epsilon[/itex]. The only thing that changes is how small [itex]\delta[/itex] must. And if we have [itex]\delta_1[/itex] such that if [itex]|x- a|<\delta_1[/itex] then [itex]|g(x)-L|< \epsilon[/itex] and, say, [itex]\delta_2[/itex] such that if [itex]|x- a|< \delta_2[/itex] then [itex]|g(x)- L|<(L^2/2)\epsilon[/itex], then taking [itex]\delta[/itex] to be the smaller of the two, if [itex]|x- a|<\delta[/itex] then both [itex]|x- a|< \delta_1[/itex] and [itex]|x- a|< \delta_2[/itex] are true and so both [itex]g(x)- L|< \epsilon[/itex] and [itex]|g(x)- L|< (L^2/2)\epsilon[/itex] are true.
     
  12. Jan 2, 2012 #11
    Ah yes, I get what you mean:) One needs to vary parameters according to the situation:) Thank you v. much!
     
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