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Homework Help: Limit Help

  1. Oct 9, 2005 #1
    This is a worked example in the book, but i cant figure out why it is this way.

    Using [tex] 0 < (x-a) < \delta [/tex] and [tex] (f(x)-L) < \epsilon [/tex] for [tex] lim_(x\rightarrow a) f(x) = L [/tex].

    Prove that [tex] lim_{x\rightarrow2} (4x-3) = 5 [/tex]

    I can understand the steps until

    [tex] {4} (x-2) < \epsilon [/tex] and [tex] 0 < (x-2) < \delta [/tex]

    then, they suddenly get, [tex] \epsilon = \delta/4 [/tex]......

    Cant really understand how they arrive at there, and they don't even show that the the limit is 5.

    I'm self taught too, so my understanding might be a little wrong.
     
    Last edited: Oct 9, 2005
  2. jcsd
  3. Oct 9, 2005 #2

    lurflurf

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    Homework Helper

    [tex]\lim_{x\rightarrow a}f(x)=L[/tex]
    means in words
    f(x) can be made as close to L as desired by making x sufficiently close to a.
    (I will use d and h instead of greak leters)
    so we want f(x) close to L
    that is we chose some h>0 and we desire that
    |f(x)-L|<h
    when x is close to a
    we make this concrete by saying
    |f(x)-L|<h
    whenever
    0<|x-a|<d(h)
    that is we look for a condition on the closeness to a that will assure f(x) is close to L. In general this condition depends upon h.
    so in the problem at hand
    f(x)=4x-3
    a=2
    L=5
    so
    |f(x)-L|=|4x-3-5|=|4x-8|=4|x-2|<h
    this is assured if
    |x-2|<h/4
    by definition (if the limit exist) the condition is expressed
    0<|x-2|<d(h)
    d(h)=h/4
    thus we have proven the given limit
    any smaller d(h) would also do
    so d(h)=h/9999999999999999
    for example would do
    the idea is just we want to bound f(x) near a
    we know f(1.999)=4.996 say
    or f(about 2)=about 5
    so we want to know for what x is 4x-3 within h of 5
    we see that it will be true when |x-2|<h/2
    or we could say
    5-|h|<f(2+h/4)<5+|h|
    so we have a relation between closeness of x to a and closeness of f(x) to L
    for example
    5-.004<f(1.999)<5+.004
     
  4. Oct 9, 2005 #3

    HallsofIvy

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    Science Advisor

    First- BE CAREFUL! It is important to state definitions precisely. There are two errors I see in:
    The first is that you don't say how those inequalities are to be used! The second is that you are missing the absolute values signs.
    The precise definition is: [tex] lim_(x\rightarrow a) f(x) = L [/tex] if and only if, for any [tex]\epsilon>0[/tex], there exist [tex]\delta> 0[/tex] such that if [tex]0<|x-a|< \delta[/tex] then [tex]|f(x)-L|< \epsilon[/tex].

    Here, a= 2, f(x)= 4x- 3, and L= 5. In order to show that that definition holds (that the limit really is 5) we must show that, for any [tex]\epsilon> 0[/tex] such a [tex]\delta[/b] really does exist. The best way to do that is to show how to find it!

    We want[\b] [tex]|f(x)-L|= \epsilon[/tex].
    But |4x-3-5|= |4x- 8|= 4|x-2| so that is the same as [tex]4|x-2|< \epsilon[/tex] or [tex]|x-2|< \frac{\epsilon}{4}[/tex].
    Now compare that with [tex]|x-2|< \delta[/tex]! Looks like a good choice for [tex]\delta[/tex] would be [tex]\frac{\epsilon}{4}[/tex] doesn't it?

    In fact, we can do exactly that. A "strict" proof would be: given [tex]\epsilon>0[/tex], let [tex]\delta= \frac{\epsilon}{4}[/tex]. Then if [tex]|x-2|< \delta[/tex], [tex]|x- 2|< \frac{\epsilon}{4}[/tex] so that [tex]4|x-2|< \epsilon[/tex].
    But then [tex]|4x- 8|= |4x-3-5|= |f(x)- L|<\epsilon[/tex] which what we needed to show.

    You are right to be a little puzzled because your book did not do it that way. Instead of starting from "let [tex]\delta= \frac{\epsilon}{4}[/tex]", as I did here, your text, like most texts, did the "preliminary part" "If I want [tex]|f(x)- L|< \epsilon[/tex], how can I find [tex]\delta[/tex]" and worked the opposite way. That is sometimes called "synthetic proof": start from what you want to show and arrive at something you know (or can make) true. Strictly speaking, a proof should go the other way- from what you know is true to what you want to prove!

    However, as long as everything you do is reversible (If you can go from [tex]4|x-2|< \epsilon[/tex] to [tex]|x-2|< \frac{\epsilon}{4}[/tex], you can certainly go from [tex]|x-2|< \frac{\epsilon}{4}[/tex] to [tex]4|x-2|< \epsilon[/tex].) then just doing it one way implies the other way.
     
    Last edited by a moderator: Oct 9, 2005
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