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Homework Help: Prove the Limit Statement

  1. Sep 15, 2009 #1


    I've just kind of been 'going through the motions,' but I feel like the Precise Definition of a Limit is about to set in.

    I think some questions about this example should help.

    Prove the Limit Statement:



    So by asserting that the limit is indeed '5' we are implying that there exists some [itex]\delta[/itex] such that for all 'x'







    [itex]\, -\epsilon<x-4<\epsilon[/itex]


    Now I am a little confused. Have I actually done anything?

    Have I shown that so long as I stay within [itex]\delta=\epsilon \text{ of }x_o[/itex] I can get within a distance of [itex]\epsilon[/itex] of 'L.'

    Because that's what i am under the impression I have done. But I am not confident about it.

    Thanks :smile:
  2. jcsd
  3. Sep 15, 2009 #2
    How about saying that 9 - x is a polynomial function and thus continuous.
  4. Sep 15, 2009 #3
    Though I can appreciate the simplicity/brevity, that is not the purpose of the exercise.

    it is to use the def'n of a limit.

  5. Sep 15, 2009 #4


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    Keep in mind that you want to show that given any [tex]\epsilon > 0[/tex] you can find a [tex]\delta[/tex] that works.

    You've done the hard part. Your argument is what might be called an exploratory argument. You started out to see how to get [tex] |(9x-5)-4 | < \epsilon[/tex] and wound up with [tex]| x-4| < \epsilon[/tex].

    Now to prove the limit statement you wish to prove that given [tex]\epsilon > 0[/tex] you can find [tex]\delta > 0[/tex] such that if [tex] 0 < |x-4| < \delta[/tex] then
    [tex] |(9x-5)-4 | < \epsilon[/tex]

    So start your final writeup like this:

    Suppose [tex]\epsilon > 0[/tex]. Let [tex]\delta = \epsilon[/tex]. (You know this is going to work from your exploratory argument, but you don't need to include that here.) Now you proceed: If [tex] 0 < |x-4| < \delta[/tex] then....now write your exploratory argument in reverse and end up with [tex] |(9x-5)-4 | < \epsilon[/tex]. That shows your [tex]\delta[/tex] works and you are done.
  6. Sep 15, 2009 #5


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    That is what is sometimes referred to as "synthetic proof". You have gone from what you want to prove, [itex]|f(x)- L|< \epsilon[/itex], to your hypothesis, [itex]|x-a|< \delta[/itex]. The "real" proof is the other way, but as long as you are sure all of yours steps are reversible you don't need to state them.

    A "real" proof would start, "Given [itex]\epsilon[/itex], choose [itex]\delta= \epsilon[/itex]. The [itex]|x- 4|< \delta= \epsilon[/itex] so that [itex]-\epsilon< x- 4< \epsilon. [itex]-\epsilon< (9- x)- 5< \epsilon[/itex], [itex]|(9- x)- 5|= |f(x)- 5|< \epsilon[/itex]

    But, as I said, as long as it is clear that all of your steps, in working from [itex]\epsilon[/itex] to [itex]delta[/itex], are reversible, you don't need to work from [itex]\delta[/itex] to [itex]\delta[/itex]. What you wrote is perfectly valid.
  7. Sep 15, 2009 #6
    OK great :smile: Thanks for the pointers. Just another quick question:

    When starting the right way, you both said to "choose epsilon=delta."

    Assume that I didn't already know that was the case. I would still make that assumption right? And if it was not the case that epsilon=delta .... what would happen?

    I get the feeling I would just end up with delta being some function of epsilon (or vice versa).

    Thanks :smile:

  8. Sep 15, 2009 #7


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    In general, epsilon=delta won't suffice (it works here basically because your function is linear with slope 1, so if you make a small change of delta in x, you've made the exact same change in y). Instead, by doing the process of starting with |f(x)-potential limit|<epsilon, you can "solve" the inequality to see what x needs to be. As an example, show that

    [tex] \lim_{x \rightarrow 0}(2x+1)=1[/tex]

    Without using any algebra of limits stuff. You'll see immediately that epsilon=delta won't suffice
  9. Sep 15, 2009 #8
    In general δ is dependent on ε and L. Also if δ suffices then any δ' where 0 < δ' < δ also suffices.

    The synthetic process mentioned earlier is usually how δ is determined and once it is found the formal proof begins with "Assume ε > 0 is given, let δ = ..." and the impliction

    [tex]0 < \left| x - a \right| < \delta \Rightarrow \left| f(x) - L \right| < \epsilon[/tex]

    is shown.

    The synthetic part is scratchwork (usually) and is frequently omitted, leading to a lot of "how'd they get that?" moments.

  10. Sep 15, 2009 #9


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    You don't "choose epsilon=delta.". epsilon is given and you choose delta = epsilon (in this problem). Delta will usually depend on epsilon and the point of your exploratory argument is to figure out a function of epsilon to set delta to make it work.
  11. Sep 16, 2009 #10
    Ah, I see :smile:

    That is a little frustrating. Thank you.
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