# Limit of a sequence 2

1. May 3, 2012

### arl146

1. The problem statement, all variables and given/known data
the sequence is an = (cos(n))^2 / 2^n

2. Relevant equations
none really

3. The attempt at a solution
like i mentioned in my last post, i usually use l'hopitals or dividing by the largest exponent from the denominator. here, i dont see why i would want to use l'hopitals, so thats out of the question. i cant really divide by the largest exponent. is there a way i could break this down or something?

2. May 3, 2012

### Curious3141

What's the maximum value the numerator can take?

3. May 3, 2012

### arl146

1. so that means the bottom is always going to be larger.
can i just write on my hw like ..

(cos(n))^2 < 1, for all n
as n approaches infinity, 2^n gets large.
small # / large # = 0

4. May 4, 2012

### Curious3141

Yes, although I would write $\cos^2 n \leq 1 \forall n \in \mathbb{N}$. The strict inequality is actually true because the only values of n that give you a value of exactly one are zero and multiples of $\pi$ and you're dealing with natural numbers here, but proving it is not worth the trouble, and is unnecessary here.

So just write that, and $\lim_{n \rightarrow \infty} 2^n = \infty$, so the quotient tends to zero.