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Math limit contradiction

  1. Jul 16, 2007 #1

    daniel_i_l

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    Gold Member

    1. The problem statement, all variables and given/known data
    You have the function f:R->R and f(x) = f(x+k) where k is in R and k>0.
    Prove or disprove:
    1) If f is continues at x0 then it's also continues at x0 + k
    2)If the limit of f at infinity is 0 then f(x)=0 for all x in R.


    2. Relevant equations



    3. The attempt at a solution
    1)Yes: If for every epsilon>0 there's a lambda>0 so that for every x where 0<|x-x0|<lambda => |f(x) - f(x0)| < epsilon then for every 0<|x-(x0+k)|<lambda => 0<|(x-k)-x0|<lambda => |f(x-k) - f(x0)| < epsilon => |f(x) - f(x0+k)| < epsilon

    2)Yes: If we have an x0 in R so that f(x0)>0 then we can choose epsilon=f(x0)/2 and then |f(x0)-0| > epsilon. So then for every N>0 we can find a k so that x0+nk > N and so |f(x0+nk)-0| > epsilon and we've proved that the limit of f at infinity isn't 0 - which is a contradiction to the information given in the question. (if there's some x0 where f(x0)<0 the proof in analogues) So for all x in R, f(x)=0.

    Are those right?
    Thanks.
     
  2. jcsd
  3. Jul 16, 2007 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Looks good to me. Congratulations.
     
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