Hi all,

http://i42.photobucket.com/albums/e316/K007/math/limofsinofalfears.gif

I'm trying to figure how to do that one, the only thing that cross my mind is something with sandwiched, any hint, thought or help about this one will be most grateful.

Is this an infinite product or nested sines? I just want to be sure what you mean.

Is it:

\lim_{n\to\infty}\frac{sin(x)\cdot sin(x)\cdot \cdot\cdot sin(x)}{n}

Or are they nested like so: sin(sin(sin(sin......sin(x)))))

I am confused by the 'n times'. Does the 'n times' represent how many times sin occurs.

Regardless, since n is the limiting factor and n\to \infty, it would appear the limit approaches 0 as n\to \infty

Is this the exact way it is presented to you?

Just making sure.

they nested sins (n represent how many times sin occurs), and I know the limit = 0, as it's easy to check this one with computer, I just don't know how to prove it.

It is just a matter of observation. X is not the limiting factor. As n gets larger and larger, it goes to 0 because of the n in the denominator.

It essentially the same as proving

\lim_{n\to \infty}\frac{x}{n} and x is a constant. Say x=1.

\lim_{n\to \infty}\frac{1}{n}=0

Maybe I mislead you or misunderstood you there is no "n" in the denominator, there is no denominator, it's :

lim (sin(sin(sin... sin(x)))), as n->inf

n it's the number of times you get this sin inside sin thing.

Well, you post seemed to me too that there was a denominator 'n' in your expression... *looks up again* yes, there is a denominator.

I will use another font next time, again there is no denominator, sorry for misleading you people, now when we cleared this up what with some help here :P.

* lim (sin(sin(sin... sin(x)))), as n->inf

n it's the number of times you get this sin inside sin thing.

This is not very rigorous, but it may suffice. By letting

L=sin(sin(sin(sin(......sin(x)))

We then note that:

sin(\underbrace{sin(sin(sin(......sin(x)))}_{ \text{this is L}})

Then we have sin(L)=L

The value that satisfies this is L=0

you say the series [x,sinx,sin(sinx),. ] is continues and monotone decreasing in [0,pi] or monotone increasing in [-pi,0] so limXn = limXn+1=L, meaning sinL=L and that only works when L=0, right?
that what you are saying?

Yes. It is monotone decreasing and bounded below. You can show the limit exists by proving this if you wish.

You may want to prove the limit sin(L)=L rigorously.

OK thanks for all the help I got my answer :P