Proving that cosine diverges as x->infinity. Not sure on how to go about this

I've to prove that as we take the limit of cosine as x goes to + infinity, the limit DNE. I've looked on different forums and sites, but all have methods I barely understand. I'm not very familiar with the Delta-Epsilon definition of limits. I've tried to look at the sticky at the top and try to understand but this is due tomorrow and I won't learn it at the moment unless it's the only way.

What's a good way to start off a proof like this? I know the reasons why it doesn't exist, as it oscillates between -1 and 1, but I need to have a proof.

Thanks

Re: Proving that cosine diverges as x->infinity. Not sure on how to go about this

Quote:

Originally Posted by

**LightningZI** I've to prove that as we take the limit of cosine as x goes to + infinity, the limit DNE. I've looked on different forums and sites, but all have methods I barely understand. I'm not very familiar with the Delta-Epsilon definition of limits. I've tried to look at the sticky at the top and try to understand but this is due tomorrow and I won't learn it at the moment unless it's the only way.

What's a good way to start off a proof like this? I know the reasons why it doesn't exist, as it oscillates between -1 and 1, but I need to have a proof.

Thanks

Are you familiar with the sequential characterization of limits?

It states that if a limit exists it is independant of the sequence used to approch the point.

Translation. For any two sequences

$\displaystyle x_n \to \infty \quad y_n \to \infty \quad f(x_n)=f(y_n)$

So if you can find two different seqences that go to infinity, but

$\displaystyle \cos(x_n) \ne \cos(y_n)$ that will show that the limit does not exist.