# Sinx and cosx diverge?

1. Jan 9, 2008

### rcmango

1. The problem statement, all variables and given/known data

why does cos x diverge?

2. Relevant equations

3. The attempt at a solution

is it because it never stops continuing to infinity? it just oscilates until 1?

and does sinx also diverge?

thanks

2. Jan 9, 2008

### mjsd

you mean they do not converge as x goes to +/- infinity?

3. Jan 9, 2008

### rcmango

yes, do they both diverge? as to, going to infinity and never stopping.

also, is this because they have an upper and lower bound, for both, as in -1 and 1 correct?

4. Jan 10, 2008

### mjsd

both functions do not converge to any particular value as x gets large. their values are bounded but they do not converge

5. Jan 10, 2008

### HallsofIvy

Do you understand why your first question made no sense? You can take the limit of a function at any value of x. You cannot talk about a limit of a function without specifying where the limit is to be taken. It is trivial that sin(x) and cox(x) converge as, say, x goes to 0 or, for that matter to any real number. Yes, both sin(x) and cos(x) diverge as x goes to infinity or -infinity. It is not because they "both have upper and lower bound". $x^2/(x^2+1)$ has "upper and lower bounds" but its limits, as x goes to either infinity or -infinity, is 0.

Do you understand what "diverge" means? It is not necessary that the value of the function go to plus or minus infinity- diverge simply means that it does not converge- that it does not, here, as x goes to infinity, get closer and closer to some specific number. Yes, both and sin(x) and cos(x) diverge (as x goes to infinity).

6. Jan 11, 2008

### Gib Z

Alot of periodic functions have the same property (im not saying all, ie f(x) = 2 is periodic for any finite interval of choice =] ) but the main reason they do not have a limit is because the value never really "hones" into a particular value, just keeps on changing. Sin x and cos x does have superior and inferior limits though (somewhat obviously) =]

7. Jan 13, 2008

### rcmango

thankyou, trying to understand series.