Is the Limit of (cos^2(t))/(t^2+1) as t Approaches Infinity Zero?

[462chevelle]

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 I am doing wrong?

 

[462chevelle]

no wonder. am i supposed to take both sides limits at infinity instead of zero?

 

[HallsofIvy]

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

 

[462chevelle]

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 don't 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

 

[HallsofIvy]

Better is 0<= cos^2(t)<= 1 like I said.

 

[462chevelle]

Better is 0<= cos^2(t)<= 1 like I said.

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