1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Prove the Limit Statement

  1. Sep 15, 2009 #1
    :smile:

    Problem

    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:

    [tex]\lim_{x\rightarrow4}(9-x)=5[/tex]

    Attempt

    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]0<|x-x_o|<\delta\Rightarrow|f(x)-5|<\epsilon[/itex]

    So:

    [itex]\-\epsilon<(9-x)-5<\epsilon[/itex]

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

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

    [itex]\therefore[/itex]

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

    [itex]|x-4|<\epsilon=\delta[/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.

    :smile:
     
  5. Sep 15, 2009 #4

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    `
    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:

    Casey
     
  8. Sep 15, 2009 #7

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.


    --Elucidus
     
  10. Sep 15, 2009 #9

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Prove the Limit Statement
  1. Prove this statement (Replies: 4)

Loading...