Limits of Sin(n)/n: Solving for N

[mjjoga]

lim (sin⁡(n)/n)=0.
The instruction say I can only use the definition of limit and no additional theorems.
So the first thing I should do is figure out if l sin(n)/n l < epsilon, find out what n is greater than. I can pull the 1/n out of the absolute value, but I don't know how to get the sine out. I mean, I know that sin(anything) would be less than say 2, but I don't know if I can do that step with the absolute value still there. Once I can define N>... I can do the rest of it.
Thanks for any help,
mjjoga

 

[Char. Limit]

Are you allowed to use geometric arguments? If you are, that is actually the best approach here.

 

[tiny-tim]

welcome to pf!

hi mjjoga! welcome to pf!

lim (sin⁡(n)/n)=0.
…
I know that sin(anything) would be less than say 2

(i assume that's the limit as n -> ∞ ?)

well, sin(anything) is ≤ 1, isn't it?

yes, if you use that the proof should be easy

 

[Char. Limit]

hi mjjoga! welcome to pf!


(i assume that's the limit as n -> ∞ ?)

well, sin(anything) is ≤ 1, isn't it?

yes, if you use that the proof should be easy

Oh, it's to infinity...

I thought it was to 0. My bad, ignore my post!

 

[tiny-tim]

he he

 

[mjjoga]

Yes, it is to infinity, my book wrote it just like that though, it's weird. I was just worried about the abs. value messing it up, but even with it it will be less than or equal to 1. Thanks guys, I appreciate it.
mjjoga