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Pn->x pn^2 -> x^2

  1. Oct 25, 2013 #1
    1. The problem statement, all variables and given/known data

    If the limit of the sequence Pn goes to x, prove that the limit of Pn^2 goes to x^2

    2. Relevant equations

    can't use epsilon and delta

    3. The attempt at a solution

    I thought simply showing that arbitrary intervals work is enough. but my teacher says that my solution doesn't make sense. any help as to my misconception would make sense.


    Let O be any open interval defined as (√a,√b) such that x is in the interval and x is the limit point of Pn. then √a < x < √b. then a < x^2 < b. Then for any open interval O1, x^2 is contained. Thus lim Pn^2 is X^2
     
  2. jcsd
  3. Oct 25, 2013 #2

    LCKurtz

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    Why don't you try looking at ##|p_n^2-x^2|## and use what you know about ##|p_n-x|##?
     
  4. Oct 25, 2013 #3
    hmmmm....

    Can I say that:

    let Pn = Sn, where the limit of Pn = x, then lim Sn = x. Then lim (Pn*Sn) = x^2. Then lim Pn^2 = x^2.


    This makes sense to me, but feels like i am cheating.
     
  5. Oct 25, 2013 #4

    Dick

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    What is Sn?? You are cheating. LCKurtz is just suggesting you use ##p_n^2-x^2=(p_n-x)(p_n+x)##.
     
  6. Oct 25, 2013 #5
    I was trying to define another sequence. I apologize for my poor form. This stuff has me totally overwhelmed.

    I thought if i defined two equivalent sequences then they would have the same limit. Then i could use like the multiplication of limits to show that it is x^2
     
  7. Oct 25, 2013 #6

    Dick

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    That's ok. I actually stand corrected. If you have the theorem that ##lim(P_n)=c## and ##lim(S_n)=d## then ##lim(P_nS_n)=cd## then that works fine. It would only be cheating if you don't have that theorem yet. Sorry.
     
  8. Oct 26, 2013 #7
    oh. I didn't know we had to prove this. It just seemed logical. I guess i will just have to go back to the drawing board. I see where you were going for in the previous post given that it factors and i can show they both have |Pn-x| but i don't know what to do with the other part since there is an addition.
     
  9. Oct 26, 2013 #8

    Dick

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    Yes, it is logical but you do have to prove it before you can take that route. But in the meantime |Pn-x| converges to zero. What does |Pn+x| converge to?
     
  10. Oct 26, 2013 #9
    Here is my shameful ignorance again. I thought that |Pn-L| meant that Pn converges to L, so |Pn-x| converges to x and |Pn+x| converges to -x. I don't see how you get zero. but that would make |Pn-x||Pn+x| converging to
    -x*x which is -x^2 ... so i know i am messing up somewhere.

    thank you for your patience btw.
     
  11. Oct 26, 2013 #10

    Dick

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    I've got all the patience in the in the world, to a certain degree. Pn converging to x means for very large n, Pn is almost equal to x. Wouldn't that mean |Pn-x| is almost zero?
     
  12. Oct 26, 2013 #11
    Yes. I see that now. So |Pn-x| = 0 then |Pn+x| = 2x? but since (Pn-x)(Pn+x) wouldn't that also go to zero? since i would essentially be having zero times 2x for very large Pn.
     
  13. Oct 26, 2013 #12

    Dick

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    |Pn-x| isn't equal to zero, but it approaches zero. And yes, |Pn+x| approaches |2x| which is a finite number. So |Pn^2-x^2| must also approach 0. So lim(Pn^2) must be what?
     
  14. Oct 26, 2013 #13
    So...
    since lim Pn goes to x as n goes to infinity,
    then
    |pn-x| goes to zero as n goes to infinity,
    then
    (Pn-x)(Pn+x) goes to (0)(2x) goes to 0 as n goes to infinity
    then
    (Pn^2 - x^2)= (Pn-x)(Pn+x) goes to 0 as n goes to infinity
    then |Pn^2-X^2| goes to 0 as n goes to infinity
    therefore the limit of Pn^2 = x^2

    did i finally get it?
     
  15. Oct 26, 2013 #14

    Dick

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    Sure you did. And I think that's probably exactly what they wanted you to do.
     
  16. Oct 26, 2013 #15
    now once i prove this in class, does that "open the door" for the use of the multiplication trick? or is that a separate rule?
     
  17. Oct 26, 2013 #16

    Dick

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    Now that is a REALLY intelligent question! I think you are getting this whole limit thing. I was just asking myself the same thing. This is a special case of the multiplication trick. It's easier to prove and maybe a little more obvious but it still needs to be proved. So it's not a separate rule. But I'd wait for it to be proved in class before you "open the door", for pedagogical reasons. Then your proof in post 3 is perfectly valid.
     
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