# Squeeze Theorem Limit

#### 462chevelle

Gold Member
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?

Related Calculus and Beyond Homework News on Phys.org

#### 462chevelle

Gold Member
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

Gold Member
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

#### HallsofIvy

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

#### 462chevelle

Gold Member
Better is 0<= cos^2(t)<= 1 like I said.
ooh sorry, i forgot the range of cos^2(t) was 0 to 1

"Squeeze Theorem Limit"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving