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Squeeze Theorem Limit

  1. Sep 18, 2014 #1

    462chevelle

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    1. The problem statement, all variables and given/known data

    Lim (cos^2(t))/(t^2+1)
    t->∞

    2. Relevant equations
    squeeze theorem -1<=Cosx<=1



    3. The attempt at a solution
    I have
    -1<=Cos(t)<=1
    (-1)^2<=Cos^2(t)<=(1)^2
    (1)/(t^2+1)<=(Cos^2(t))/(t^2+1)<=(1)/(t^2+1)
    I took both of limits of the 2 outsides as t->0
    i got -1 and 1. so the limit should not exist. But i think this is incorrect. Any hints on what im doing wrong?
     
  2. jcsd
  3. Sep 18, 2014 #2

    462chevelle

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    no wonder. am i supposed to take both sides limits at infinity instead of zero?
     
  4. Sep 18, 2014 #3

    HallsofIvy

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    Uhh, yes, the problem says "[itex]t\to \infty[/itex]", not 0. But you don't want to say "[itex]-1\le cos(t)\le 1[/itex] therefore [itex]1\le cos(t)\le 1[/itex]"! If x is somewhere between -1 and 1 then [itex]x^2[/itex] is somewhere between 0 and 1, NOT "between 1 and 1"! Draw a graph of [itex]y= x^2[/itex] for [itex]-1\le x\le 1[/itex] to see that.
     
  5. Sep 18, 2014 #4

    462chevelle

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    oh, ok. so when i put cos like
    -1<=Cos(t)<=1 i should start out like
    -1<=cos^2(t)<=1.
    that way i dont have to square the cos then get from 1<cos<1

    so there i get the limit of each is 0 so the limit of the entire function must be 0
     
  6. Sep 19, 2014 #5

    HallsofIvy

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    Better is 0<= cos^2(t)<= 1 like I said.
     
  7. Sep 19, 2014 #6

    462chevelle

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    ooh sorry, i forgot the range of cos^2(t) was 0 to 1
     
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