Question

What is the limit of `cosx` as x goes to infinity?

Answer 1

There is no limit.

Explanation:

Recall or Note:

`lim_(xrarroo)f(x) = L` if and only if

for every positived `epsilon`, there is an `M` that satisfies:
for all `x > M`, `abs(f(x) - L) < epsilon`

As `x` increases without bound, `cosx` continues to attain every value between `-1` and `1`. So it cannot be getting and staying within `epsilon` of some one number, `L`,

Answer 2

Refer to explanation

Explanation:

Choose two sequences such as

`a_n=2n*pi` and `b_n=(2n+1)*pi` where

`lim_(n->oo) a_n=oo` and `lim_(n->oo) b_n=oo`

But `lim_(n->oo)cos(a_n)=cos(2pin)=1` and `lim_(n->oo)cos(b_n)=cos((2n+1)pi)=-1`

Hence there is no limit for `cosx` as `x->oo`

The theorem that was used for the proof is

**(Divergence Criterion for Functional Limits):
Let `f:A→R` `f:A→R`, and let `c` be a limit point of `A`. If there exist two sequences `(x_n)` and `(y_n)` in `A` with `x_n≠c` and `y_n≠c`, and:
`lim_(n->oo)x_n=lim_(n->oo) y_n=c` but `lim_(n->oo)f(x_n)≠lim_(n->oo)f(y_n)`
then we can conclude that the functional `lim_(x→c)f(x)` does not exist.