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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


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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
    when x is close to a
    we make this concrete by saying
    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
    this is assured if
    by definition (if the limit exist) the condition is expressed
    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
    so we have a relation between closeness of x to a and closeness of f(x) to L
    for example
  4. Oct 9, 2005 #3


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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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