Homework Help: 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.