Squeeze Theorem Limit

  • #1
462chevelle
Gold Member
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Homework Statement



Lim (cos^2(t))/(t^2+1)
t->∞

Homework Equations


squeeze theorem -1<=Cosx<=1



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?
 

Answers and Replies

  • #2
462chevelle
Gold Member
305
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no wonder. am i supposed to take both sides limits at infinity instead of zero?
 
  • #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.
 
  • #4
462chevelle
Gold Member
305
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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
 
  • #5
HallsofIvy
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Better is 0<= cos^2(t)<= 1 like I said.
 
  • #6
462chevelle
Gold Member
305
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Better is 0<= cos^2(t)<= 1 like I said.

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

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